JEE Main 2025 Jan 22 Shift 2, Mathematics Q6: Vector (or Cross) Product of Two Vectors

Parametrise $Q$ on the line through $P$ in the given direction: $Q=P+\lambda (3,2,2)=(-2+3\lambda ,-1+2\lambda ,3+2\lambda )$, with $\lambda \ne 0$ so $Q\ne P$.

$\overrightarrow{QR}=R-Q=(3-3\lambda ,4-2\lambda ,-2\lambda )$.

$|QR{|}^2=25$:

$(3-3\lambda {)}^2+(4-2\lambda {)}^2+4{\lambda }^2=25$

$9-18\lambda +9{\lambda }^2+16-16\lambda +4{\lambda }^2+4{\lambda }^2=25$

$17{\lambda }^2-34\lambda +25=25\Rightarrow 17{\lambda }^2-34\lambda =0\Rightarrow \lambda =2$ (rejecting $\lambda =0$).

So $Q=(4,3,7)$. Then $\overrightarrow{PQ}=(6,4,4)$ and $\overrightarrow{PR}=(3,4,0)$.

$\overrightarrow{PQ}\times \overrightarrow{PR}=\left|\begin{array}{c}\hat{i}\hat{j}\hat{k}\\ 644\\ 340\end{array}\right|=\hat{i}(0-16)-\hat{j}(0-12)+\hat{k}(24-12)=-16\hat{i}+12\hat{j}+12\hat{k}$.

$|\overrightarrow{PQ}\times \overrightarrow{PR}{|}^2=256+144+144=544$.

Area${}^2=\frac{1}{4}|\overrightarrow{PQ}\times \overrightarrow{PR}{|}^2=\frac{544}{4}=136$.