Let $P(3, 2, 6)$ be a point in space and $Q$ be a point on the line $\vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(-3\hat{i} + \hat{j} + 5\hat{k})$. Then the value of $\mu$ for which the vector $\overrightarrow{PQ}$ is parallel to the plane $x - 4y + 3z = 1$ is

$\text{Given: } P(3, 2, 6)$

$Q \in \vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(-3\hat{i} + \hat{j} + 5\hat{k})$

$\text{Plane: } x - 4y + 3z = 1$

$\text{To find: } \mu \text{ such that } \overrightarrow{PQ} \parallel \text{ plane } x-4y+3z=1$

$\text{Plan: } PQ \parallel \text{plane} \iff \overrightarrow{PQ} \perp \vec{n}$

$\vec{n} = (1, -4, 3)$

$\text{Set } \overrightarrow{PQ} \cdot \vec{n} = 0$

$Q = (1-3\mu,\ -1+\mu,\ 2+5\mu)$

$\overrightarrow{PQ} = Q - P = (1-3\mu-3,\ -1+\mu-2,\ 2+5\mu-6)$

$= (-2-3\mu,\ -3+\mu,\ -4+5\mu)$

$\overrightarrow{PQ} \cdot \vec{n} = 0$

$(-2-3\mu)(1) + (-3+\mu)(-4) + (-4+5\mu)(3) = 0$

$-2 - 3\mu + 12 - 4\mu - 12 + 15\mu = 0$

$8\mu - 2 = 0 \Rightarrow \mu = \dfrac{1}{4}$

$Q = \left(\dfrac{1}{4},\ -\dfrac{3}{4},\ \dfrac{13}{4}\right)$

$\overrightarrow{PQ} = \left(-\dfrac{11}{4},\ -\dfrac{11}{4},\ -\dfrac{11}{4}\right)$

$\overrightarrow{PQ} \cdot \vec{n} = -\dfrac{11}{4} + 11 - \dfrac{33}{4} = 0\ \checkmark$

$\boxed{\mu = \dfrac{1}{4}}$

$\text{Answer: (1)}$