Let PQR be a triangle such that $$\overrightarrow{PQ}=-2\widehat{i}-\widehat{j}+2\widehat{k}$$ and $$\overrightarrow{PR}=a\widehat{i}+b\widehat{j}-4\widehat{k},a,b \in Z$$. Let S be the point on QR, which is equidistant from the lines PQ and PR. If $$|\overrightarrow{PR}|=9$$ and $$\overrightarrow{PS}=\widehat{i}-7\widehat{j}+2\widehat{k}$$, then the value of 3a - 4b is_______

$$\vec{PQ} = -2\hat{i} - \hat{j} + 2\hat{k} \implies \vert{}\vec{PQ}\vert{} = \sqrt{(-2)^2 + (-1)^2 + 2^2} = 3$$

$$\vec{PR} = a\hat{i} + b\hat{j} - 4\hat{k} \implies \vert{}\vec{PR}\vert{} = \sqrt{a^2 + b^2 + 16} = 9 \implies a^2 + b^2 = 65 \quad \text{--- (1)}$$

$$\vec{PS} = \hat{i} - 7\hat{j} + 2\hat{k} \implies \vert{}\vec{PS}\vert{} = \sqrt{1^2 + (-7)^2 + 2^2} = \sqrt{54} = 3\sqrt{6}$$

$$\cos\theta = \frac{\vec{PQ} \cdot \vec{PS}}{\vert{}\vec{PQ}\vert{}\vert{}\vec{PS}\vert{}} = \frac{(-2)(1) + (-1)(-7) + (2)(2)}{3 \cdot 3\sqrt{6}} = \frac{-2 + 7 + 4}{9\sqrt{6}} = \frac{9}{9\sqrt{6}} = \frac{1}{\sqrt{6}}$$

Using the angle bisector condition $$\cos\theta = \frac{\vec{PS} \cdot \vec{PR}}{\vert{}\vec{PS}\vert{}\vert{}\vec{PR}\vert{}}$$:

$$\frac{1}{\sqrt{6}} = \frac{(1)(a) + (-7)(b) + (2)(-4)}{3\sqrt{6} \cdot 9}$$

$$\frac{1}{\sqrt{6}} = \frac{a - 7b - 8}{27\sqrt{6}} \implies a - 7b - 8 = 27 \implies a - 7b = 35 \quad \text{--- (2)}$$

$$a = 35 + 7b$$

$$(35 + 7b)^2 + b^2 = 65 \implies 1225 + 490b + 50b^2 = 65$$

$$50b^2 + 490b + 1160 = 0 \implies 5b^2 + 49b + 116 = 0$$

$$(5b + 29)(b + 4) = 0 \implies b = -4 \quad (\text{since } b \in \mathbb{Z})$$

$$a = 35 + 7(-4) = 7$$

$$3a - 4b = 3(7) - 4(-4) = 21 + 16 = 37$$