Let P Q R be a triangle such that overrightarrowP Q=-2 i- j+2 k and overrightarrow PR=a i+b j-4 k, a, b ∈ mathbbZ. Let S be the point on QR, which is…

Given: $\vec{PQ} = -2\hat{i} - \hat{j} + 2\hat{k}$PQ​=−2i^−j^​+2k^, $\vec{PR} = a\hat{i} + b\hat{j} - 4\hat{k}$PR=ai^+bj^​−4k^ ($a, b \in \mathbb{Z}$a,b∈Z), $\vec{PS} = \hat{i} - 7\hat{j} + 2\hat{k}$PS=i^−7j^​+2k^, $|\vec{PR}| = 9$∣PR∣=9

From $|\vec{PR}| = 9$∣PR∣=9: $a^2 + b^2 + 16 = 81 \Rightarrow a^2 + b^2 = 65$a2+b2+16=81⇒a2+b2=65 ...(1)

Since S is on QR equidistant from lines PQ and PR, PS is the angle bisector. So:

$$\cos\theta = \frac{\vec{PQ} \cdot \vec{PS}}{|\vec{PQ}||\vec{PS}|} = \frac{-2+7+4}{3 \cdot 3\sqrt{6}} = \frac{9}{9\sqrt{6}} = \frac{1}{\sqrt{6}}$$cosθ=∣PQ​∣∣PS∣PQ​⋅PS​=3⋅36​−2+7+4​=96​9​=6​1​

Similarly: $\frac{1}{\sqrt{6}} = \frac{\vec{PS} \cdot \vec{PR}}{|\vec{PS}||\vec{PR}|} = \frac{a - 7b - 8}{3\sqrt{6} \cdot 9}$6​1​=∣PS∣∣PR∣PS⋅PR​=36​⋅9a−7b−8​

$a - 7b = 35$a−7b=35 ...(2)

From (1) and (2): $a = 7, b = -4$a=7,b=−4

$3a - 4b = 21 + 16 = 37$3a−4b=21+16=37 (NTA Answer)

However, $\vec{QS} = 3\hat{i} - 6\hat{j}$QS​=3i^−6j^​ and $\vec{SR} = 6\hat{i} + 3\hat{j} - 6\hat{k}$SR=6i^+3j^​−6k^ are not parallel, so Q, S, R are not collinear. Also $\cos\theta = \frac{2\sqrt{2}}{\sqrt{3}} > 1$cosθ=3​22​​>1, which is impossible. Hence no such triangle exists.

So, this should be a bonus question.