The distance of the point $(1,-2,3)$ from the plane $x-y+z=5$ measured parallel to a line,whose direction ratios are $2,3,-6$ is :

The line $\frac{x - 3}{2} = \frac{y - 4}{3} = \frac{z - 5}{4}$ lies in the plane $4x + 4y - kz - d = 0$. The values of $k$ and $d$ are

The equation of the plane passing through the point $(2,-1,-3)$ and parallel to the lines $\frac{x-1}{3}=\frac{y+2}{2}=\frac{z}{-4}$ and $\frac{x}{2}=\frac{y-1}{-3}=\frac{z-2}{2}$ is

The distance between the line $\frac{x - 1}{3} = \frac{y + 2}{-2} = \frac{z - 1}{2}$ and the plane $2x + 2y - z = 6$ is

The point of intersection of the line $\frac{x - 1}{3} = \frac{y + 2}{4} = \frac{z - 3}{-2}$ and the plane $2x - y + 3z - 1 = 0$ is:

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 $\vec{PQ}$ is parallel to the plane $x - 4y + 3z = 1$ is