If the perpendicular distance of the plane passing through the point $Q(1, 0, -1)$ and containing the line $\vec{r} = (\hat{i} - 3\hat{j} + \hat{k}) + \lambda(2\hat{i} - 2\hat{j} + \hat{k})$ from the origin is $\frac{p}{\sqrt{53}}$, then $p = $

The vector equation of a plane in parametric form, passing through the points $A(-1, 2, 0)$ and $B(2, 2, -1)$ and parallel to the line $\frac{x - 1}{1} = \frac{2y + 1}{2} = \frac{z + 1}{-1}$ is:

The angle between the line $\vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(3\hat{i}+\hat{j})$ and the plane $\vec{r} \cdot (\hat{i}+2\hat{j}+3\hat{k})=8$ is:

Let the line $\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}$ intersect the lines $\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}$ and $\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}$ at the points $A$ and $B$ respectively. Then the distance of the mid-point of the line segment $AB$ from the plane $2x-2y+z=14$ is

Find the coordinates of the point where the line passing through $(3, -4, -5)$ and $(2, -3, 1)$ intersects the plane $2x + y + z = 7$.

The distance of the point $(2, 3, -5)$ from the plane $x + 2y - 2z = 9$ is