JEE Advanced 2025 — Paper 1 — Question 19

Let $L_{1}: \vec{r}=\vec{a}+t \vec{b}$

$\vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 1 & 2 & 1\end{array}\right|=\hat{i}(1)-\hat{j}(1)+\hat{k}(1)$

Dr's of $\mathrm{L}_{1}:<1,-1,1>$

$\mathrm{L}_{1}: \frac{\mathrm{x}+1}{1}=\frac{\mathrm{y}-0}{-1}=\frac{\mathrm{z}-6}{1}$

$\mathrm{L}_{2}: \frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}+1}{-1}=\frac{\mathrm{z}-3}{1}=\lambda$

M : $2 \mathrm{x}+\mathrm{y}-2 \mathrm{z}-6=0$

Let point on $\mathrm{L}_{2}(\lambda+2,-\lambda-1, \lambda+3)$

$2(\lambda+2)+(-\lambda-1)-2 (\lambda+3)-6=0$

$2 \lambda+4-3 \lambda-13=0$

$\lambda=-9$

$\therefore \mathrm{Q}(-7,8,-6)$

Line PR: $\frac{x-2}{2}=\frac{y+1}{1}=\frac{z-3}{-2}=\mu$

$\mathrm{R}(2 \mu+2, \mu-1,-2 \mu+3)$

Put in plane $2(2 \mu+2)+(\mu-1)-2(-2 \mu+3)-6=0$

$4 \mu+4+\mu-1+4 \mu-6-6=0$

$9 \mu-9=0 \Rightarrow \mu=1$

$\mathrm{R}(4,0,1)$\ $\mathrm{P}(2,-1,3) \& \mathrm{Q}(-7,8,-6)$

$\mathrm{PQ}=\sqrt{81+81+81}=9 \sqrt{3}$

$\mathrm{Q}(-7,8,-6)$ & $\mathrm{R}(4,0,1)$

$\mathrm{QR}=\sqrt{121+64+49}=\sqrt{234}$

Area ( $\triangle \mathrm{PQR}$ )

$=\frac{1}{2}|\overrightarrow{\mathrm{QP}} \times \overrightarrow{\mathrm{QR}}|$

$=\frac{1}{2}\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 9 & -9 & 9 \\ 11 & -8 & 7\end{array}\right|$

$=\frac{3}{2} \sqrt{234}$

$\overrightarrow{\mathrm{PQ}}=-9 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}-9 \hat{\mathrm{k}}=-9(\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}})$

$\overrightarrow{\mathrm{PR}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-2 \hat{\mathrm{k}}$

$\cos \theta=\left|\frac{\overrightarrow{\mathrm{PQ}} \cdot \overrightarrow{\mathrm{PR}}}{\mathrm{PQ} \cdot \mathrm{PR}}\right|$

$=\frac{9}{9 \sqrt{3} \times 3}=\frac{1}{3 \sqrt{3}}$

$\theta=\cos ^{-1}\left(\frac{1}{3 \sqrt{3}}\right)$