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

If the line joining the points $\hat{i}+\hat{j}$ and $3 \hat{i}+\hat{j}-\hat{k}$ meets the plane that passes through the point $2 \hat{i}+4 \hat{j}$ and is parallel to the vectors $3 \hat{j}+5 \hat{k}$ and $3 \hat{i}-\hat{k}$ at point $P$,then the position vector of the point $P$ is

The distance of the point $(-1, 2, -2)$ from the line of intersection of the planes $2x + 3y + 2z = 0$ and $x - 2y + z = 0$ is:

The angle between the line with the direction ratios $(2, 5, 1)$ and the plane $8x + 2y - z = 4$ is

The length of the projection of the line segment joining the points $(5, -1, 4)$ and $(4, -1, 3)$ on the plane $x + y + z = 7$ is:

If the equation of the line passing through the point $(0, -\frac{1}{2}, 0)$ and perpendicular to the lines $\overrightarrow{r} = \lambda(\hat{i} + a\hat{j} + b\hat{k})$ and $\overrightarrow{r} = (\hat{i} - \hat{j} - 6\hat{k}) + \mu(-b\hat{i} + a\hat{j} + 5\hat{k})$ is $\frac{x-1}{-2} = \frac{y+4}{d} = \frac{z-c}{-4}$,then $a+b+c+d$ is equal to :