shahkb4
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Solution
assuming all oranges are identical
Let children $\mathrm{C}_{1}, \mathrm{C}_{2}, \mathrm{C}_{3}, \mathrm{C}_{4}$ get number of oranges $\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}, \mathrm{x}_{4}$
$\mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}+\mathrm{x}_{4}=16$ where $\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}, \mathrm{x}_{4} \geq 1$
let $\mathrm{x}_{\mathrm{i}}=\mathrm{x}_{\mathrm{i}}^{\prime}+1 ; \mathrm{x}_{1}^{\prime}+\mathrm{x}_{2}^{\prime}+\mathrm{x}_{3}^{\prime}+\mathrm{x}_{4}^{\prime}=12$
Number of non-negative integral solutions ${ }^{n+p-1} C_{p-1}={ }^{12+4-1} C_{4-1}$
$={ }^{15} \mathrm{C}_{3}=455$ Answer(4)Solution
$(\mathrm{P}, \mathrm{P}, \mathrm{P})(\mathrm{QQ})(\mathrm{RR}) \mathrm{S}, \mathrm{T}, \mathrm{U}, \mathrm{V}$
(i) 3 same, 1 diff. $(\mathrm{P}, \mathrm{P}, \mathrm{P}, \mathrm{R}) \rightarrow 1 \times{ }^{6} \mathrm{C}_{1} \times\left(\frac{4!}{3!}\right)=6 \times 4=24$
(ii) 2 same, 2 diff. ( $\mathrm{Q}, \mathrm{Q}, \mathrm{P}, \mathrm{T}) \rightarrow{ }^{3} \mathrm{C}_{1} \times{ }^{6} \mathrm{C}_{2} \times\left(\frac{4!}{2!}\right)=3 \times 15 \times 12=450+90=540$
(iii) 2 same, 2 same $(\mathrm{P}, \mathrm{P}, \mathrm{R}, \mathrm{R}) \rightarrow{ }^{3} \mathrm{C}_{2} \times\left(\frac{4!}{2!2!}\right)=\frac{3 \times 24}{4}=18$
(iv) 4 different $(\mathrm{P}, \mathrm{R}, \mathrm{U}, \mathrm{V}) \rightarrow{ }^{7} \mathrm{C}_{4}(4!)=\frac{7.6 .5}{6} \times 24=42(20)=840$Total such words $=24+540+18+840=1422$ Answer(1422)
Solution
First Method:
$\mathrm{S}=\{1,2,3,4, \ldots, 10,11\}$
even no $\rightarrow 2,4,6,8,10$
odd no $\rightarrow 1,3,5,7,9,11$
Total subsets $=2^{11}$
No of subsets such that product of numbers is odd $=2^{6}$
No of subsets such that product of number is even $=2^{11}-2^{6}=2048-64=1984$ required no of subsets $=1984-(\mathrm{n}(\mathrm{B})=1$ \& Product is even)\[ =1984-5=1979 \text { Ans. } \]Second Method:
\[ \begin{array}{ll} \mathrm{n}(\mathrm{~B})=2, & { }^{11} \mathrm{C}_{2}-{ }^{6} \mathrm{C}_{2}(\text { all are odds }) \\ \mathrm{n}(\mathrm{~B})=3, & { }^{11} \mathrm{C}_{3}-{ }^{6} \mathrm{C}_{3}(\text { all are odds }) \\ \mathrm{n}(\mathrm{~B})=4, & { }^{11} \mathrm{C}_{4}-{ }^{6} \mathrm{C}_{4}(\text { all are odds }) \\ \mathrm{n}(\mathrm{~B})=5, & { }^{11} \mathrm{C}_{5}-{ }^{6} \mathrm{C}_{5}(\text { all are odds }) \\ \mathrm{n}(\mathrm{~B})=6, & { }^{11} \mathrm{C}_{6}-{ }^{6} \mathrm{C}_{6}(\text { all are odds }) \\ \mathrm{n}(\mathrm{~B})=7, & { }^{11} \mathrm{C}_{7}-0 \\ \mathrm{n}(\mathrm{~B})=8 & { }^{11} \mathrm{C}_{8} \\ \mathrm{n}(\mathrm{~B})=9, & { }^{11} \mathrm{C}_{9} \\ \mathrm{n}(\mathrm{~B})=10, & { }^{11} \mathrm{C}_{10} \\ \mathrm{n}(\mathrm{~B})=11, & { }^{11} \mathrm{C}_{11} \end{array} \]required no. of subsets $=\sum_{\mathrm{r}=2}^{11}{ }^{11} \mathrm{C}_{\mathrm{r}}-\sum_{\mathrm{r}=2}^{6}{ }^{6} \mathrm{C}_{\mathrm{r}}$
$=\left(2^{11}-{ }^{11} \mathrm{C}_{0}-{ }^{11} \mathrm{C}_{1}\right)-\left(2^{6}-{ }^{6} \mathrm{C}_{0}-{ }^{6} \mathrm{C}_{1}\right)$
$=2048-1-11-(64-1-6)$
$=2048-64-5=1979$ Answer(1979)Solution
Required no. of pentagons $=660$ Answer(660)
Solution
$6^{\mathrm{m}}+9^{\mathrm{n}}=(5+1)^{\mathrm{m}}+(10-1)^{\mathrm{n}}$
$=\left({ }^{m} \mathrm{C}_{0} 5^{m}+\ldots+{ }^{m} \mathrm{C}_{m-1} 5+{ }^{m} \mathrm{C}_{m}\right)+\left({ }^{n} \mathrm{C}_{0} 10^{n}-{ }^{n} \mathrm{C}_{1} 10^{n-1}+\cdots+{ }^{n} \mathrm{C}_{n-1} 10+{ }^{n} \mathrm{C}_{n}(-1)^{n}\right]$
$=(5 \mathrm{k}+1)+\left(10 \lambda+(-1)^{\mathrm{n}}\right)$
$=5(\mathrm{k}+2 \lambda)+1+(-1)^{\mathrm{n}}$
$m \in\{1,2,3,4, \ldots, 49,50\}$
n must be an odd no i.e. $\mathrm{n} \in\{1,3,5,7, \ldots, 49\}$;
p $=50 \times 25=1250$
$\mathrm{m}+\mathrm{n}=2^{2}=4$
$(\mathrm{m}, \mathrm{n})=(1,3),(2,2),(3,1) \rightarrow 3$
$m+n=3^{2}=9$
$(\mathrm{m}, \mathrm{n})=(1,8),(2,7)(3,6), \ldots(8,1) \rightarrow 8$
$\mathrm{m}+\mathrm{n}=5^{2}=25$
$(\mathrm{m}, \mathrm{n})=(1,24)(2,23),(3,21) \ldots(24,1) \rightarrow 24$
$\mathrm{m}+\mathrm{n}=7^{2}=49$
$(m, m)=(1,48)(2,47),(3,46), \ldots(48,1) \rightarrow 48$
$(\mathrm{m}+\mathrm{n})_{\text {max }}=100$
$\mathrm{m}+\mathrm{n} \neq 11^{2}=121$
$\mathrm{q}=3+8+24+48=83$
$\therefore p+q=1250+83=1333$ Answer(1333)Solution
line $\mathrm{L}_{1}: \frac{\mathrm{x}-4}{2}=\frac{\mathrm{y}-0}{0}=\frac{\mathrm{z}+1}{3}=\lambda$
line $\mathrm{L}_{2}: \frac{\mathrm{x}-43}{3}=\frac{\mathrm{y}-\alpha}{-1}=\frac{\mathrm{z}-\beta}{0}=\mu$
$\mathrm{Q}(2 \lambda+4,0,3 \lambda-1) \equiv \mathrm{Q}(3 \mu+43,-\mu+\alpha, \beta)$
$2 \lambda+4=3 \mu+43 \Rightarrow 2 \lambda=3 \mu+39$
$0=-\mu+\alpha=\alpha=\mu$\begin{figure}
\captionsetup{labelformat=empty}
\caption{[JEE MAIN 28 Jan 2026 S II]}
\includegraphics[alt={},max width=\textwidth]{https://cdn.mathpix.com/cropped/27a95272-36b0-4ba6-b451-34902f9c19d8-17.jpg?height=351&width=530&top_left_y=339&top_left_x=910}
\end{figure}
$3 \lambda-1=\beta$
$\mathrm{Q}(3 \alpha+43,0, \beta) \mathrm{P}(43, \alpha, \beta)$
$\mathrm{PQ}=13 \sqrt{10} \Rightarrow \sqrt{9 \alpha^{2}+\alpha^{2}+0}=13 \sqrt{10}$
$\Rightarrow|\alpha| \sqrt{10}=13 \sqrt{10} \Rightarrow \alpha= \pm 13$
Case I : $\alpha=13, \mu=13$, given $\beta<0$
From (1), $2 \lambda=3 \mu+39=78 \Rightarrow \lambda=39$
From (2), $\beta=3 \lambda-1=3 \times 39-1>0$ (rejected)
Case II : $\alpha=-13, \mu=-13$
From (1), $2 \lambda=-39+39 \Rightarrow \lambda=0$
From (3), $\beta=3 \lambda-1=-1$
Now $\alpha^{2}+\beta^{2}=169+1=170$ Answer(170)Solution
line $\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}=\lambda$
$\mathrm{M}(\lambda+1,2 \lambda, \lambda+1)$
$\mathrm{PM} \xrightarrow{\mathrm{drs}} \lambda-2,2 \lambda-2, \lambda$
line $\xrightarrow{\text { drs }} 1,2,1$
$\mathrm{PM} \perp$ line $\Rightarrow \lambda-2+4 \lambda-4+\lambda=0$
$\Rightarrow 6 \lambda=6 \Rightarrow \lambda=1$
we get $\mathrm{M}(2,2,2)$ and $\mathrm{Q}(1,2,3)$
Now distance of Q from another line
we get $\mathrm{M}(2,2,2)$ and $\mathrm{Q}(1,2,3)$
Now distance of Q from another line\[ \begin{aligned} & \frac{\mathrm{x}-9}{3}=\frac{\mathrm{y}-9}{2}=\frac{\mathrm{z}-5}{-2}=\mathrm{t} \\ & \mathrm{R}(3 \mathrm{t}+9,2 \mathrm{t}+9,-2 \mathrm{t}+5) \\ & \mathrm{QR} \xrightarrow{\mathrm{drs}} 3 \mathrm{t}+8,2 \mathrm{t}+7,-2 \mathrm{t}+2 \\ & \text { line } \xrightarrow{\mathrm{drs}} 3,2,-2 \end{aligned} \]$\mathrm{QR} \perp$ line
$(3 t+8) 3+2(2 t+7)-2(-2 t+2)=0$
$\Rightarrow 17 \mathrm{t}+24+14-4=0$
$\Rightarrow \mathrm{t}=-2, \mathrm{R}(3,5,9)$
$\mathrm{QR} \equiv \sqrt{4+9+36}=7$Second Method:
$\overrightarrow{\mathrm{AQ}}=-8 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}$
$\overrightarrow{\operatorname{AR}} \|(3 \hat{i}+2 \hat{j}-2 \hat{k})$
$|\overrightarrow{\mathrm{QR}}|=|\overrightarrow{\mathrm{AQ}}| \sin \theta=\frac{|\overrightarrow{\mathrm{AQ}} \times \overrightarrow{\mathrm{AR}}|}{|\overrightarrow{\mathrm{AR}}|}=|\overrightarrow{\mathrm{AQ}} \times \mathrm{AR}|$
$=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ -8 & -7 & -2 \\ \frac{3}{\sqrt{17}} & \frac{2}{\sqrt{17}} & -\frac{2}{\sqrt{17}}\end{array}\right|=\frac{1}{\sqrt{17}}|18 \hat{i}-22 \hat{j}+5 \hat{k}|$
$=\sqrt{\frac{324+484+25}{17}}=\sqrt{\frac{833}{17}}=\sqrt{49}=7$Third Method
$\overrightarrow{\mathrm{AQ}}=-8 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-2 \hat{\mathrm{k}} ; \mathrm{AQ}=\sqrt{64+49+4}=\sqrt{117}$
$\mathrm{AR}=$ Projection of $\overrightarrow{\mathrm{AQ}}$ on given line whose $\operatorname{drs} 3,2,-2$
$=\left|\frac{-8(3)-7(2)-2(-2)}{\sqrt{9+4+4}}\right|=\frac{|-24-14+4|}{\sqrt{17}}=2 \sqrt{17}$
$\mathrm{QR}=\sqrt{\mathrm{AQ}^{2}-\mathrm{AR}^{2}}=\sqrt{117-4 \times 17}=\sqrt{49}=7$ Answer(3)Solution
Point $\mathrm{B}(3 \lambda+1,4 \lambda+2, \mathrm{~b} \lambda+\mathrm{a})$ on line $\mathrm{L}_{1}$
It also lies on given line L
$\frac{3 \lambda+1-1}{1}=\frac{4 \lambda+2}{2}=\frac{\mathrm{b} \lambda+\mathrm{a}-1}{1}$
$\Rightarrow 3 \lambda=2 \lambda+1=\mathrm{b} \lambda+\mathrm{a}-1$
$\Rightarrow \lambda=1,3=\mathrm{b}+\mathrm{a}-1 \Rightarrow \mathrm{a}+\mathrm{b}=4$
$B(4,6, a+b) \equiv B(4,6,4)$
similarly Point C lies on $\mathrm{L}_{2}$
$c(\mu+1,4 \mu+2, c \mu+a)$
It lies on line L also
$\frac{\mu+1-1}{1}=\frac{4 \mu+2}{2}=\frac{\mathrm{c} \mu+\mathrm{a}-1}{1}$
$\Rightarrow \mu=2 \mu+1=\mathrm{c} \mu+\mathrm{a}-1$
$\Rightarrow \mu=-1$ and $\mathrm{a}=\mathrm{c}$
$\mathrm{C}(0,-2, \mathrm{a}-\mathrm{c}) \equiv \mathrm{C}(0,-2,0)$
$\because \mathrm{AB}=\mathrm{AC} \Rightarrow \mathrm{AB}^{2}=\mathrm{AC}^{2}$
$\Rightarrow 3^{2}+4^{2}+(\mathrm{a}-4)^{2}=1^{2}+4^{2}+\mathrm{a}^{2}$
$\Rightarrow 8+\mathrm{a}^{2}-8 \mathrm{a}+16=\mathrm{a}^{2}$
$\Rightarrow 89=24 \Rightarrow \mathrm{a}=3=\mathrm{c}, \mathrm{b}=1$
$\therefore \mathrm{a}+\mathrm{b}+\mathrm{c}=7$ Answer(1)Solution
First line $\mathrm{L}_{1}$ passes through point $\mathrm{A}(-1,2,4)$ with drs $\alpha,-1,-\alpha$
First line $\mathrm{L}_{1}$ passes through point $\mathrm{B}(0,1,1)$ with drs $\alpha, 2,2 \alpha$
shortest distance $\mathrm{b} / \mathrm{w}$ skew lines $=\frac{\left|\begin{array}{ccc}\mathrm{x}_{2}-\mathrm{x}_{1} & \mathrm{y}_{2}-\mathrm{y}_{1} & \mathrm{z}_{2}-\mathrm{z}_{1} \\ \mathrm{a}_{1} & \mathrm{~b}_{1} & \mathrm{c}_{1} \\ \mathrm{a}_{2} & \mathrm{~b}_{2} & \mathrm{c}_{2}\end{array}\right|}{\left\|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ \mathrm{a}_{1} & \mathrm{~b}_{1} & \mathrm{c}_{1} \\ \mathrm{a}_{2} & \mathrm{~b}_{2} & \mathrm{c}_{2}\end{array}\right\|}$
$=\frac{\left|\begin{array}{ccc}1 & -1 & -3 \\ \alpha & -1 & -\alpha \\ \alpha & 2 & 2 \alpha\end{array}\right|}{\left\|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ \alpha & -1 & -\alpha \\ \alpha & 2 & 2 \alpha\end{array}\right\|}=\sqrt{2}$
$\Rightarrow \frac{\left|(-2 \alpha+2 \alpha)+1\left(2 \alpha^{2}+\alpha^{2}\right)-3(3 \alpha)\right|}{\left|0-3 \alpha^{2} \hat{\mathrm{j}}+3 \alpha \hat{\mathrm{k}}\right|}=\sqrt{2}$
$\Rightarrow \frac{\left|3 \alpha^{2}-9 \alpha\right|}{\sqrt{9 \alpha^{4}+9 \alpha^{2}}}=\sqrt{2} \Rightarrow \frac{3|\alpha-3||\alpha|}{3 \alpha \sqrt{\alpha^{2}+1}}=\sqrt{2}$
$\Rightarrow|\alpha-3|=\sqrt{2} \sqrt{\alpha^{2}+1} \Rightarrow \alpha^{2}+9-6 \alpha=2 \alpha^{2}+2$
$\Rightarrow \alpha^{2}+6 \alpha-7=0 \Rightarrow(\alpha+7)(\alpha-1)=0$
$\alpha=-7,1$ sum of $\alpha=-6$ Answer(2)Solution
$\mathrm{L}_{1} \xrightarrow{\text { drs }} 4,1,1, \mathrm{~L}_{2} \xrightarrow{\text { drs }} 1,1,0$
required $\mathrm{L}, \mathrm{L} \perp \mathrm{L}_{1}$ \& $\mathrm{L} \perp \mathrm{L}_{2}$
$\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 4 & 1 & 1 \\ 1 & 1 & 0\end{array}\right|=-\hat{i}+\hat{j}+3 \hat{j}$
line $\mathrm{L} \xrightarrow{\text { drs }}-1,1,3$
line $\mathrm{L} \frac{\mathrm{x}-1}{-1}=\frac{\mathrm{y}-1}{1}=\frac{\mathrm{z}-1}{3}=\lambda \mathrm{Q}(-\lambda+1, \lambda+1,3 \lambda+1)$
Q lies on yz plane
$\mathrm{Q}_{\mathrm{x}}=0 \Rightarrow-\lambda+1=0 \Rightarrow \lambda=1$
$\mathrm{Q}(0,2,4) \mathrm{S}(1,0,-1)$
$\overrightarrow{\mathrm{PQ}}=-\hat{i}+\hat{j}+3 \hat{k}, \overrightarrow{\mathrm{PS}}=-\hat{j}-2 \hat{k}$
area of parallelogram $=|\overrightarrow{\mathrm{PQ}} \times \overrightarrow{\mathrm{PS}}|=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ -1 & 1 & 3 \\ 0 & 1 & -2\end{array}\right|$\[ =|\hat{\imath}-2 \hat{j}+\hat{k}|=\sqrt{6} \text { Answer(6) } \]Solution
$\mathrm{R}(2 \lambda+1,3 \lambda+2,4 \lambda+3)$ on $\mathrm{L}_{1}$
$\mathrm{R}(5 \mu+4,2 \mu+1, \mu)$ on $\mathrm{L}_{2}$
$2 \lambda+1=5 \mu+4 \mid 3 \lambda+2=2 \mu+1$
$\begin{aligned} &=\lambda=\frac{5 \mu}{2}+\frac{3}{2} \Rightarrow \frac{15 \mu}{2}+\frac{9}{2}+1=2 \mu \\ & \Rightarrow \mu=-1, \lambda=-1, \mathrm{R}(-1-1,-1)\end{aligned}$
$\mathrm{P}(2 \lambda+1,3 \lambda+2,4 \lambda+3)$ on $\mathrm{L}_{1}$
$\mathrm{Q}(5 \mu+4,2 \mu+1, \mu)$ on $\mathrm{L}_{2}$
$|\overrightarrow{\mathrm{PR}}|^{2}=29$
$\Rightarrow(2 \lambda+2)^{2}+(3 \lambda+3)^{2}+(4 \lambda+4)^{2}=29$
$\Rightarrow 29 \lambda^{2}+4+9+16+8 \lambda+18 \lambda+32 \lambda=29$
$\Rightarrow 29 \lambda^{2}+58 \lambda=0 \Rightarrow 29 \lambda(\lambda+2)=0$
$\Rightarrow \lambda=0, \mathrm{P}(1,2,3)$ (accepted)
or $\lambda=-2, \mathrm{P}(-3,-4,-5)$ (rejected)
as P lies in First octant
$|\overrightarrow{\mathrm{PQ}}|^{2}=(5 \mu+3)^{2}+(2 \mu-1)^{2}+(\mu-3)^{2}=\frac{47}{3}$\[ \begin{aligned} & \Rightarrow 25 \mu^{2}+9+30 \mu+4 \mu^{2}+1-4 \mu+\mu^{2}+9-6 \mu=\frac{47}{3} \\ & \Rightarrow 30 \mu^{2}+20 \mu+\frac{10}{3}=0 \Rightarrow(3 \mu+1)^{2}=0 \Rightarrow \mu=-\frac{1}{3} \\ & \mathrm{Q}\left(\frac{7}{3}, \frac{1}{3},-\frac{1}{3}\right) \mathrm{R}(-1,-1,-1) \\ & \begin{array}{l} 27(\mathrm{QR})^{2}=27\left(\left(\frac{10}{3}\right)^{2}+\left(\frac{4}{3}\right)^{2}+\left(+\frac{2}{3}\right)^{2}\right) \\ \quad=3(120)=360 \text { Answer }(2) \end{array} \end{aligned} \]Solution
$\mathrm{L}_{1}: \frac{\mathrm{x}}{2}=\frac{\mathrm{y}+\mathrm{a}}{1}=\frac{\mathrm{z}}{1}=\lambda$
$\mathrm{A}(2 \lambda, \lambda-\mathrm{a}, \lambda)$
$\mathrm{L}_{2}: \frac{\mathrm{x}-2 \mathrm{~b}}{2}=\frac{\mathrm{y}-\mathrm{a}}{2}=\frac{\mathrm{z}+2 \mathrm{~b}}{-5}=\mu$
$\mathrm{A}(2 \mu+2 \mathrm{~b}, \mu+\mathrm{a},-5 \mu-2 \mathrm{~b})$
$\Rightarrow \lambda=\mu+\mathrm{b}, \lambda=\mu+2 \mathrm{a}, \mu+2 \mathrm{a}=-5 \mu-2 \mathrm{~b}$
From first two, $\mathrm{b}=2 \mathrm{a}$
From third, $6 \mu=-2 b-2 a=-4 a-2 a$
$\lambda=\mathrm{a}, \mu=-\mathrm{a}$
We have Point $\mathrm{A}(2 \mathrm{a}, 0, \mathrm{a})$ and $\mathrm{P}(\mathrm{a}, 2, \mathrm{a})$
$\mathrm{AP} \xrightarrow{\text { drs }} \mathrm{a},-2,0 ; \mathrm{L}_{1} \xrightarrow{\text { drs }} 2,1,1$
$\mathrm{AP} \perp \mathrm{L}_{1}$
$2 \mathrm{a}-2+0=0 \Rightarrow \mathrm{a}=1$
$\mathrm{b}=2 \mathrm{a}=2 \therefore \mathrm{a}+\mathrm{b}=3$ Answer(3)Solution
$4 \ell+\mathrm{m}-\mathrm{n}=0 \Rightarrow \mathrm{n}=4 \ell+\mathrm{m}$
$\mathrm{n}(2 \mathrm{~m}+10 \ell)+3 \ell \mathrm{~m}=0$
$\Rightarrow(4 \ell+\mathrm{m})(2 \mathrm{~m}+10 \ell)+3 \ell \mathrm{~m}=0$
$\Rightarrow 8 \ell \mathrm{~m}+40 \ell^{2}+2 \mathrm{~m}^{2}+10 \ell \mathrm{~m}+3 \ell \mathrm{~m}=0$
$\Rightarrow 40 \ell^{2}+16 \ell \mathrm{~m}+5 \ell \mathrm{~m}+2 \mathrm{~m}^{2}=0$
$\Rightarrow 8 \ell(5 \ell+2 \mathrm{~m})+\mathrm{m}(5 \ell+2 \mathrm{~m})=0$
$\Rightarrow(5 \ell+2 \mathrm{~m})(8 \ell+\mathrm{m})=0$
$5 \mathrm{l}+2 \mathrm{~m}=0 \quad \& \quad 8 \ell+\mathrm{m}=0$
$\Rightarrow \mathrm{m}=-\frac{5 \ell}{2}$
$\Rightarrow \mathrm{m}=-8 \ell$
$\mathrm{n}=4 \ell+\mathrm{m}$
$\mathrm{n}=4 \ell+\mathrm{m}=4 \ell-8 \ell$
$\mathrm{n}=4 \ell-\frac{5 \ell}{2}=\frac{3 \ell}{2}$
$\Rightarrow \mathrm{n}=-4 \ell$
$\ell, \mathrm{m}, \mathrm{m} \rightarrow \ell,-\frac{5 \ell}{2}, \frac{3 \ell}{2}$
$\ell, \mathrm{m}, \mathrm{n} \rightarrow \ell,-8 \ell,-4 \ell$
or line $\mathrm{L}_{1} \mathrm{drs} 2,-5,3 \mid$ line $\mathrm{L}_{2} \mathrm{drs}-1,8,4$\[ \cos \theta=\left|\frac{2(-1)-5(8)+3(4)}{\sqrt{4+25+9} \sqrt{1+64+16}}\right|=\frac{30}{\sqrt{38} \times 9}=\frac{10}{3 \sqrt{38}} \text { Answer }(4) \]Solution
First Method:
line $\mathrm{L}: \frac{\mathrm{x}}{1}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}-2}{3}=\lambda$
$\mathrm{D}(\lambda, 2 \lambda+1, \lambda \lambda+2)$\[ \mathrm{AD} \xrightarrow{\mathrm{drs}} \lambda-1,2 \lambda+1-6,3 \lambda+2-3 \]
or $\lambda-1,2 \lambda-5,3 \lambda-1$\[ \mathrm{BC} \xrightarrow{\text { drs }} 1,2,3 \]\[ \mathrm{AD} \perp \mathrm{BC} \Rightarrow(\lambda-1) 1+(2 \lambda-5) 2+3(3 \lambda-1)=0 \Rightarrow 14 \lambda=14 \Rightarrow(\lambda=1) \]\[ \mathrm{D}(1,3,5) \mathrm{A}(1,6,3) \]\[ \mathrm{AD}=\sqrt{0+3^{2}+2^{2}}=\sqrt{13} \]\[ \Delta \mathrm{ABC}=\frac{1}{2}(\mathrm{BC}) \mathrm{AD}=\frac{1}{2}(10) \sqrt{13}=5 \sqrt{13} \]Second Method:
$\mathrm{B}(4,9, \alpha)$ lies on line BC
$\frac{4}{1}=\frac{9-1}{2}=\frac{\alpha-2}{3} \Rightarrow \alpha=14$
$\mathrm{B}(4,9,14) ; \mathrm{A}(1,6,3)$
$\overrightarrow{\mathrm{AB}}=3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+11 \hat{\mathrm{k}} ; \overrightarrow{\mathrm{BC}}=10 \frac{(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})}{\sqrt{1+4+9}}$
$\cos \theta=\frac{\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{BC}}}{|\overrightarrow{\mathrm{AB}} \| \overrightarrow{\mathrm{BC}}|}=\frac{10}{\sqrt{14}} \frac{(3+6+33)}{10 \sqrt{9+9+121}}$
$\cos \theta=\frac{3 \sqrt{14}}{\sqrt{139}} ; \sin \theta=\sqrt{1-\cos ^{2} \theta}=\frac{\sqrt{13}}{\sqrt{139}}$
$\Delta \mathrm{ABC}=\frac{1}{2}(\mathrm{AB})(\mathrm{BC}) \sin \theta=\frac{1}{2}|3 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+11 \hat{\mathrm{k}}| 10\left(\frac{\sqrt{13}}{\sqrt{139}}\right)$
$=5 \sqrt{13} \sqrt{9+9+121}\left(\frac{1}{\sqrt{139}}\right)=5 \sqrt{13}$Third Method:
$\overrightarrow{\mathrm{AB}}=3 \hat{i}+3 \hat{j}+11 \hat{k} ; \overrightarrow{\mathrm{BC}}=\frac{10}{\sqrt{14}}(\hat{i}+2 \hat{j}+3 \hat{k})$
$\Delta \mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathrm{BA}} \times \overrightarrow{\mathrm{BC}}|=\frac{1}{2} \frac{10}{\sqrt{14}}\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 3 & 3 & 11 \\ 1 & 2 & 3\end{array}\right|$
$=\frac{5}{\sqrt{14}}|-13 \hat{i}+2 \hat{j}+3 \hat{k}|=\frac{5}{\sqrt{14}} \sqrt{169+4+9}=\frac{5 \sqrt{182}}{\sqrt{14}}=5 \sqrt{13}$ Answer(1)Solution
line $\mathrm{L} \equiv \frac{\mathrm{x}+1}{2}=\frac{\mathrm{y}+1}{3}=\frac{\mathrm{z}+3}{6}=\lambda=\mathrm{t}$
$\mathrm{A}(2 \lambda-1,3 \lambda-1,6 \lambda-3)$
$\mathrm{L}_{1}: \frac{\mathrm{x}+1}{2}=\frac{\mathrm{y}+1}{3}=\frac{\mathrm{z}-9}{0}=\mu$
$\mathrm{A}(2 \mu-1,3 \mu-1,9)$
$2 \lambda-1=2 \mu-1 \Rightarrow \lambda=\mu$
$6 \lambda-3=9 \Rightarrow 6 \lambda=12 \Rightarrow \lambda=2$
A(3,5,9)
Point $\mathrm{P}(2 \mathrm{t}-1,3 \mathrm{t}-1,6 \mathrm{t}-3)$
$\mathrm{AP}=7$
$\Rightarrow(2 \mathrm{t}-4)^{2}+(3 \mathrm{t}-6)^{2}+(6 \mathrm{t}-12)^{2}=49$
$\Rightarrow(\mathrm{t}-1)(\mathrm{t}-3)=0 \Rightarrow \mathrm{t}=1,3$
$\left.\begin{array}{ll}\mathrm{t}=1 & \mathrm{P}_{1}(1,2,3) \\ \mathrm{t}=3 & \mathrm{P}_{2}(5,8,15)\end{array}\right]$
$\sum_{\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{S}}(\mathrm{a}+\mathrm{b}+\mathrm{c})=1+2+3+5+8+15=34$ Answer $(1)$
