[shahkb4]

shahkb4

3 June, 2026 at 8:15 pm #1077

Let the lines $\mathrm{L}_{1}: \overrightarrow{\mathrm{r}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \mathrm{k}+\lambda(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \mathrm{k}), \lambda \in \mathrm{R}$ and
$L_{2}: \vec{r}=(4 \hat{i}+\hat{j})+\mu(5 \hat{i}+2 \hat{j}+k), \mu \in R$, intersect at the point $R$.
Let P and Q be the points lying on lines $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ respectively, such that $|\overrightarrow{\mathrm{PR}}|=\sqrt{29}$ and $|\overrightarrow{\mathrm{PQ}}|=\sqrt{\frac{47}{3}}$. If the point P lies in the first octant, then $27(\mathrm{QR})^{2}$ is equal to
[JEE MAIN 24 Jan 2026 S I]
(1) 340
(2) 360
(3) 320
(4) 348

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[shahkb4]

shahkb4

3 June, 2026 at 8:15 pm #1078

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
$|\overrightarrow{\mathrm{PQ}}|^{2}=(5 \mu+3)^{2}+(2 \mu-1)^{2}+(\mu-3)^{2}=\frac{47}{3}$

\[ \begin{aligned} & \Rightarrow 25 \mu^{2}+9+30 \mu+4 \mu^{2}+1-4 \mu+\mu^{2}+9-6 \mu=\frac{47}{3} \\ & \Rightarrow 30 \mu^{2}+20 \mu+\frac{10}{3}=0 \Rightarrow(3 \mu+1)^{2}=0 \Rightarrow \mu=-\frac{1}{3} \\ & \mathrm{Q}\left(\frac{7}{3}, \frac{1}{3},-\frac{1}{3}\right) \mathrm{R}(-1,-1,-1) \\ & \begin{array}{l} 27(\mathrm{QR})^{2}=27\left(\left(\frac{10}{3}\right)^{2}+\left(\frac{4}{3}\right)^{2}+\left(+\frac{2}{3}\right)^{2}\right) \\ \quad=3(120)=360 \text { Answer }(2) \end{array} \end{aligned} \]

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