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

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

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

$\Rightarrow\ Q = (1-3\mu,\ -1+\mu,\ 2+5\mu) \text{ lies on the line}$

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

$\Leftrightarrow\ \overrightarrow{PQ} \cdot \vec{n} = 0,\quad \vec{n} = (1, -4, 3)$

$\vec{b} = -3\hat{i} + \hat{j} + 5\hat{k}\quad (\text{direction of line})$

$\vec{n} = \hat{i} - 4\hat{j} + 3\hat{k}\quad (\text{normal of plane})$

$\overrightarrow{PQ} = Q - P = (-2-3\mu,\ -3+\mu,\ -4+5\mu)$

$\text{Plan: } \overrightarrow{PQ} \cdot \vec{n} = 0 \Rightarrow \text{solve for } \mu$

$\overrightarrow{PQ} = (-2-3\mu,\ -3+\mu,\ -4+5\mu),\quad \vec{n} = (1,-4,3)$

$\overrightarrow{PQ}\cdot\vec{n} = 1(-2-3\mu) + (-4)(-3+\mu) + 3(-4+5\mu)$

$= (-2-3\mu) + (12-4\mu) + (-12+15\mu)$

$= -2 + 8\mu$

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

$Q = \left(\dfrac14,\ -\dfrac34,\ \dfrac{13}{4}\right)$

$\text{At } \mu=\dfrac14:\ \overrightarrow{PQ} = -\dfrac{11}{4}(1,1,1)$

$(1,1,1)\cdot(1,-4,3) = 1-4+3 = 0 \ \checkmark$

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

$\boxed{\text{option } (1)}$