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ABBC CA 2 2 $1 \rightarrow{ }^{4} \mathrm{C}_{2}{ }^{5} \mathrm{C}_{2}{ }^{4} \mathrm{C}_{1}=6 \times 10 \times 4=240$ 2 1 $2 \rightarrow{ }^{4} \mathrm{C}_{2}{ }^{5} \mathrm{C}_{1}{ }^{4} \mathrm{C}_{2}=6(5) 6=180$ 1 2 $2 \rightarrow{ }^{4} \mathrm{C}_{1}{ }^{5} \mathrm{C}_{2}{ }^{4} \mathrm{C}_{2}=4(10) 6=240$ 
Required no. of pentagons = 660 Answer(660)Solution
line $\mathrm{L}_{1}: \frac{\mathrm{x}-4}{2}=\frac{\mathrm{y}-0}{0}=\frac{\mathrm{z}+1}{3}=\lambda$
line $\mathrm{L}_{2}: \frac{\mathrm{x}-43}{3}=\frac{\mathrm{y}-\alpha}{-1}=\frac{\mathrm{z}-\beta}{0}=\mu$
$\mathrm{Q}(2 \lambda+4,0,3 \lambda-1) \equiv \mathrm{Q}(3 \mu+43,-\mu+\alpha, \beta)$\[ 2 \lambda+4=3 \mu+43 \Rightarrow 2 \lambda=3 \mu+39 \]\[ 0=-\mu+\alpha=\alpha=\mu \]
\[ 3 \lambda-1=\beta \]\[ \mathrm{Q}(3 \alpha+43,0, \beta) \mathrm{P}(43, \alpha, \beta) \]\[ \begin{aligned} & P Q=13 \sqrt{10} \Rightarrow \sqrt{9 \alpha^{2}+\alpha^{2}+0}=13 \sqrt{10} \\ & \Rightarrow|\alpha| \sqrt{10}=13 \sqrt{10} \Rightarrow \alpha= \pm 13 \end{aligned} \]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 $\frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}-1}{1}=\lambda$
$\mathrm{M}(\lambda+1,2 \lambda, \lambda+1)$
$\mathrm{PM} \xrightarrow{\mathrm{drs}} \lambda-2,2 \lambda-2, \lambda$
line $\xrightarrow{\mathrm{drs}} 1,2,1$
$\mathrm{PM} \perp$ 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\[ \begin{aligned} & \frac{x-9}{3}=\frac{y-9}{2}=\frac{z-5}{-2}=t \\ & R(3 t+9,2 t+9,-2 t+5) \\ & Q R \xrightarrow{d r s} 3 t+8,2 t+7,-2 t+2 \end{aligned} \]
line $\xrightarrow{\mathrm{drs}} 3,2,-2$
QR ⟂ line\[ \begin{aligned} & (3 \mathrm{t}+8) 3+2(2 \mathrm{t}+7)-2(-2 \mathrm{t}+2)=0 \\ & \Rightarrow 17 \mathrm{t}+24+14-4=0 \\ & \Rightarrow \mathrm{t}=-2, \mathrm{R}(3,5,9) \\ & \mathrm{QR} \equiv \sqrt{4+9+36}=7 \end{aligned} \]Second Method:
\[ \begin{aligned} & \overrightarrow{\mathrm{AQ}}=-8 \hat{i}-7 \hat{j}-2 \hat{k} \\ & \overrightarrow{\mathrm{AR}} \|(3 \hat{i}+2 \hat{j}-2 \hat{k}) \end{aligned} \]
\[ \begin{aligned} & |\overrightarrow{\mathrm{QR}}|=|\overrightarrow{\mathrm{AQ}}| \sin \theta=\frac{|\overrightarrow{\mathrm{AQ}} \times \overrightarrow{\mathrm{AR}}|}{|\overrightarrow{\mathrm{AR}}|}=\mid \overrightarrow{\mathrm{AQ}} \times \mathrm{AR} \\ & =\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ -8 & -7 & -2 \\ \frac{3}{\sqrt{17}} & \frac{2}{\sqrt{17}} & -\frac{2}{\sqrt{17}} \end{array}\right|=\frac{1}{\sqrt{17}}|18 \hat{\mathrm{i}}-22 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}| \\ & =\sqrt{\frac{324+484+25}{17}}=\sqrt{\frac{833}{17}}=\sqrt{49}=7 \end{aligned} \]Third Method
\[ \begin{aligned} & \overrightarrow{\mathrm{AQ}}=-8 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}-2 \hat{\mathrm{k}} ; \mathrm{AQ}=\sqrt{64+49+4}=\sqrt{117} \\ & \mathrm{AR}=\text { Projection of } \overrightarrow{\mathrm{AQ}} \text { on given line whose drs } 3,2,-2 \\ & =\left|\frac{-8(3)-7(2)-2(-2)}{\sqrt{9+4+4}}\right|=\frac{|-24-14+4|}{\sqrt{17}}=2 \sqrt{17} \\ & \mathrm{QR}=\sqrt{\mathrm{AQ}^{2}-\mathrm{AR}^{2}}=\sqrt{117-4 \times 17}=\sqrt{49}=7 \text { Answer }(3) \end{aligned} \]
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
\[ \begin{aligned} & \frac{3 \lambda+1-1}{1}=\frac{4 \lambda+2}{2}=\frac{b \lambda+a-1}{1} \\ & \Rightarrow 3 \lambda=2 \lambda+1=b \lambda+a-1 \\ & \Rightarrow \lambda=1,3=b+a-1 \Rightarrow a+b=4 \\ & B(4,6, a+b) \equiv B(4,6,4) \end{aligned} \]
similarly Point C lies on $\mathrm{L}_{2}$\[ c(\mu+1,4 \mu+2, c \mu+a) \]It lies on line L also
\[ \begin{aligned} & \frac{\mu+1-1}{1}=\frac{4 \mu+2}{2}=\frac{c \mu+a-1}{1} \\ & \Rightarrow \mu=2 \mu+1=c \mu+a-1 \\ & \Rightarrow \mu=-1 \text { and } a=c \\ & C(0,-2, a-c) \equiv C(0,-2,0) \\ & \because A B=A C \Rightarrow A B^{2}=A C^{2} \\ & \Rightarrow 3^{2}+4^{2}+(a-4)^{2}=1^{2}+4^{2}+a^{2} \\ & \Rightarrow 8+a^{2}-8 a+16=a^{2} \\ & \Rightarrow 89=24 \Rightarrow a=3=c, b=1 \\ & \therefore a+b+c=7 \text { Answer }(1) \end{aligned} \]
