JEE Main 2026 — Vector Algebra Question with Solution

Given: $\overrightarrow{PQ}=-2\hat{i}-\hat{j}+2\hat{k}$, $\overrightarrow{PR}=a\hat{i}+b\hat{j}-4\hat{k}$ ($a,b\in \mathbb{Z}$), $\overrightarrow{PS}=\hat{i}-7\hat{j}+2\hat{k}$, $|\overrightarrow{PR}|=9$
From $|\overrightarrow{PR}|=9$: $a^2+b^2+16=81\Rightarrow a^2+b^2=65$ ...(1)
Since S is on QR equidistant from lines PQ and PR, PS is the angle bisector. So:
$\cos \theta =\frac{\overrightarrow{PQ}\cdot \overrightarrow{PS}}{|\overrightarrow{PQ}||\overrightarrow{PS}|}=\frac{-2+7+4}{3\cdot 3\sqrt{6}}=\frac{9}{9\sqrt{6}}=\frac{1}{\sqrt{6}}$
Similarly: $\frac{1}{\sqrt{6}}=\frac{\overrightarrow{PS}\cdot \overrightarrow{PR}}{|\overrightarrow{PS}||\overrightarrow{PR}|}=\frac{a-7b-8}{3\sqrt{6}\cdot 9}$
$a-7b=35$ ...(2)
From (1) and (2): $a=7,b=-4$
$3a-4b=21+16=37$ (NTA Answer)
However, $\overrightarrow{QS}=3\hat{i}-6\hat{j}$ and $\overrightarrow{SR}=6\hat{i}+3\hat{j}-6\hat{k}$ are not parallel, so Q, S, R are not collinear. Also $\cos \theta =\frac{2\sqrt{2}}{\sqrt{3}}>1$, which is impossible. Hence no such triangle exists.
So, this should be a bonus question.