Find Parameter for Vector Parallel to Given Plane

To find the value of $\mu$ for which the vector $\vec{PQ}$ is parallel to the given plane, we proceed as follows:

The coordinates of the point $P$ are given as:
$P = (3, 2, 6)$

A point $Q$ lies on the line $\vec{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(-3\hat{i} + \hat{j} + 5\hat{k})$. Hence, the position vector of $Q$ in terms of the parameter $\mu$ is:
$\vec{r}_Q = (1 - 3\mu)\hat{i} + (-1 + \mu)\hat{j} + (2 + 5\mu)\hat{k}$

The vector $\vec{PQ}$ is obtained by subtracting the position vector of $P$ from the position vector of $Q$:
$\vec{PQ} = \vec{r}_Q - \vec{r}_P = \left[(1 - 3\mu) - 3\right]\hat{i} + \left[(-1 + \mu) - 2\right]\hat{j} + \left[(2 + 5\mu) - 6\right]\hat{k}$
$\vec{PQ} = (-2 - 3\mu)\hat{i} + (\mu - 3)\hat{j} + (5\mu - 4)\hat{k}$

The equation of the given plane is:
$x - 4y + 3z = 1$

The normal vector to this plane is:
$\vec{n} = \hat{i} - 4\hat{j} + 3\hat{k}$

For the vector $\vec{PQ}$ to be parallel to the plane, it must be perpendicular to the normal vector $\vec{n}$ of the plane:
$\vec{PQ} \cdot \vec{n} = 0$

Taking the scalar (dot) product:
$(-2 - 3\mu)(1) + (\mu - 3)(-4) + (5\mu - 4)(3) = 0$

Expanding the expression:
$-2 - 3\mu - 4\mu + 12 + 15\mu - 12 = 0$

Combining like terms:
$8\mu - 2 = 0$
$8\mu = 2 \implies \mu = \frac{1}{4}$

Thus, the value of $\mu$ is $\frac{1}{4}$.

Therefore, the correct option is A.