Problem Analysis: We are given a triangle PQR with vectors $\vec{PQ}$ and $\vec{PR}$. A point S on QR is equidistant from lines PQ and PR. We are given $|\vec{PR}| = 9$ and $\vec{PS}$. We need to find $3a - 4b$, where $\vec{PR} = a\hat{i} + b\hat{j} - 4\hat{k}$ and $a, b \in \mathbb{Z}$.

Step 1: Calculate Magnitude of $\vec{PQ}$

Given $\vec{PQ} = -2\hat{i} - \hat{j} + 2\hat{k}$.

The magnitude is:

$|\vec{PQ}| = \sqrt{(-2)^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$

Step 2: Use Magnitude of $\vec{PR}$

Given $\vec{PR} = a\hat{i} + b\hat{j} - 4\hat{k}$ and $|\vec{PR}| = 9$.

Squaring the magnitude:

$|\vec{PR}|^2 = a^2 + b^2 + (-4)^2 = 9^2$

$a^2 + b^2 + 16 = 81$

$a^2 + b^2 = 65$

Since $a, b$ are integers, possible pairs $(a, b)$ are $(\pm 1, \pm 8)$, $(\pm 8, \pm 1)$, $(\pm 4, \pm 7)$, $(\pm 7, \pm 4)$.

Step 3: Apply Equidistant Condition

The condition that S is equidistant from lines PQ and PR means the distance from S to line PQ equals the distance from S to line PR.

The distance from point S to a line passing through P with direction vector $\vec{D}$ is given by $\frac{|\vec{PS} \times \vec{D}|}{|\vec{D}|}$.

So, $\frac{|\vec{PS} \times \vec{PQ}|}{|\vec{PQ}|} = \frac{|\vec{PS} \times \vec{PR}|}{|\vec{PR}|}$.

Substituting the known magnitudes:

$\frac{|\vec{PS} \times \vec{PQ}|}{3} = \frac{|\vec{PS} \times \vec{PR}|}{9}$

Multiplying by 9 gives:

$3 |\vec{PS} \times \vec{PQ}| = |\vec{PS} \times \vec{PR}|$

Step 4: Calculate Cross Products

Calculate $\vec{PS} \times \vec{PQ}$:

$\vec{PS} \times \vec{PQ} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -7 & 2 \\ -2 & -1 & 2 \end{vmatrix}$

$= \hat{i}((-7)(2) - (2)(-1)) - \hat{j}((1)(2) - (2)(-2)) + \hat{k}((1)(-1) - (-7)(-2))$

$= \hat{i}(-14 + 2) - \hat{j}(2 + 4) + \hat{k}(-1 - 14)$

$= -12\hat{i} - 6\hat{j} - 15\hat{k}$

$|\vec{PS} \times \vec{PQ}| = \sqrt{(-12)^2 + (-6)^2 + (-15)^2} = \sqrt{144 + 36 + 225} = \sqrt{405} = 9\sqrt{5}$

Calculate $\vec{PS} \times \vec{PR}$:

$\vec{PS} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -7 & 2 \\ a & b & -4 \end{vmatrix}$

$= \hat{i}((-7)(-4) - (2)(b)) - \hat{j}((1)(-4) - (2)(a)) + \hat{k}((1)(b) - (-7)(a))$

$= \hat{i}(28 - 2b) - \hat{j}(-4 - 2a) + \hat{k}(b + 7a)$

$= (28 - 2b)\hat{i} + (4 + 2a)\hat{j} + (b + 7a)\hat{k}$

$|\vec{PS} \times \vec{PR}|^2 = (28 - 2b)^2 + (4 + 2a)^2 + (b + 7a)^2$

$|\vec{PS} \times \vec{PR}|^2 = 4(14 - b)^2 + 4(2 + a)^2 + (7a + b)^2$

Step 5: Substitute and Solve for $a, b$

Substitute the magnitudes back into the equation from Step 3:

$3 (9\sqrt{5}) = |\vec{PS} \times \vec{PR}|$

$27\sqrt{5} = |\vec{PS} \times \vec{PR}|$

Squaring both sides:

$(27\sqrt{5})^2 = |\vec{PS} \times \vec{PR}|^2$

$729 \times 5 = 4(14 - b)^2 + 4(2 + a)^2 + (7a + b)^2$

$3645 = 4(196 - 28b + b^2) + 4(4 + 4a + a^2) + (49a^2 + 14ab + b^2)$

$3645 = 784 - 112b + 4b^2 + 16 + 16a + 4a^2 + 49a^2 + 14ab + b^2$

$3645 = 53a^2 + 5b^2 + 16a - 112b + 14ab + 800$

$2845 = 53a^2 + 5b^2 + 16a - 112b + 14ab$

Step 6: Test Integer Pairs

We use the condition $a^2 + b^2 = 65$ and test the integer pairs.

Let's test the pair $(a, b) = (7, -4)$:

Check $a^2 + b^2 = 7^2 + (-4)^2 = 49 + 16 = 65$. (Satisfied)

Substitute into the equation from Step 5:

$53(7)^2 + 5(-4)^2 + 16(7) - 112(-4) + 14(7)(-4)$

$= 53(49) + 5(16) + 112 + 448 - 392$

$= 2597 + 80 + 112 + 448 - 392$

$= 3237 - 392 = 2845$

This pair satisfies the equation. Thus, $a = 7$ and $b = -4$.

Step 7: Calculate the Final Value

We need to find $3a - 4b$.

$3a - 4b = 3(7) - 4(-4)$

$= 21 + 16 = 37$