Let P Q R be a triangle such that P Q =-2 i - j +2 k and PR =a i +b...

Given: $\vec{PQ} = -2\hat{i} - \hat{j} + 2\hat{k}$, $\vec{PR} = a\hat{i} + b\hat{j} - 4\hat{k}$ ($a, b \in \mathbb{Z}$), $\vec{PS} = \hat{i} - 7\hat{j} + 2\hat{k}$, $|\vec{PR}| = 9$
From $|\vec{PR}| = 9$: $a^2 + b^2 + 16 = 81 \Rightarrow a^2 + b^2 = 65$ ...(1)
Since S is on QR equidistant from lines PQ and PR, PS is the angle bisector. So:
$\cos\theta = \frac{\vec{PQ} \cdot \vec{PS}}{|\vec{PQ}||\vec{PS}|} = \frac{-2+7+4}{3 \cdot 3\sqrt{6}} = \frac{9}{9\sqrt{6}} = \frac{1}{\sqrt{6}}$
Similarly: $\frac{1}{\sqrt{6}} = \frac{\vec{PS} \cdot \vec{PR}}{|\vec{PS}||\vec{PR}|} = \frac{a - 7b - 8}{3\sqrt{6} \cdot 9}$
$a - 7b = 35$ ...(2)
From (1) and (2): $a = 7, b = -4$
$3a - 4b = 21 + 16 = 37$ (NTA Answer)
However, $\vec{QS} = 3\hat{i} - 6\hat{j}$ and $\vec{SR} = 6\hat{i} + 3\hat{j} - 6\hat{k}$ 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.