Let P(3, 2, 6) be a point in space and θ 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 $\vec{PQ}$ is parallel to the plane x - 4y + 3z = 1, is

$\frac{1}{4}$

Let the position vector of Q be

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

$= (-3\mu + 1) \hat{i} + (\mu - 1) \hat{j} + (5\mu + 2) \hat{k}.$

Then , $\vec{PQ}= (-3\mu - 2) \hat{i} + (\mu - 3) \hat{j} + (5 \mu - 4) \hat{k}$

It is given that $\vec{PQ}$ is parallel to the plane x - 4y + 3z = 1 whose normal is $\vec{n} = \hat{i} - 4\hat{j} + 3\hat{k}$.

$∴ \vec{PQ}.\vec{n} = 0 $

$⇒ (-3\mu -2) - 4(\mu - 3) + 3 (5\mu - 4) = 0 ⇒ \mu = \frac{1}{4}$