JEE Main 2025 — 22 January, Evening Shift — Question 6

$\overrightarrow{\mathrm{PQ}}$ parallel to $3 \hat{i}+2 \hat{j}+2 \hat{k}, R(1,3,3)$

$\Rightarrow \mathrm{Q}(3 \lambda-2,2 \lambda-1,2 \lambda+3), \lambda \in \mathrm{R}-\{0\}$

$|\overrightarrow{\mathrm{QR}}|=5=\sqrt{(3 \lambda-3)^{2}+(2 \lambda-4)^{2}+(2 \lambda)^{2}}$

$\therefore 17 \lambda^{2}-34 \lambda+25=25 \Rightarrow \lambda=2(\because \lambda \neq 0)$

$\therefore \mathrm{Q}(4,3,7), \mathrm{P}(-2,-1,3), \mathrm{R}(1,3,3)$

Area of $\triangle \mathrm{PQR}=[\mathrm{PQR}]=\frac{1}{2}|\overrightarrow{\mathrm{PQ}} \times \overrightarrow{\mathrm{PR}}|$

$[\mathrm{PQR}]=\frac{1}{2}\left\|\begin{array}{lll}\hat{i} & \hat{\mathrm{j}} & \hat{k} \\6 & 4 & 4 \\3 & 4 & 0\end{array}\right\|=\left\|\begin{array}{lll}\hat{i} & \hat{\mathrm{j}} & \hat{k}\\ 3 & 2 & 2 \\3 & 4 & 0\end{array}\right\|$

$[\mathrm{PQR}]=|-8 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}|=\sqrt{136}$

$\therefore[\mathrm{PQR}]^{2}=136$