The distance of the point $(1, -2, 4)$ from the plane passing through the point $(1, 2, 2)$ and perpendicular to the planes $x - y + 2z = 3$ and $2x - 2y + z + 12 = 0$ is:

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

The distance of the point $P(1, 0, -3)$ from the plane $x - y - z = 9$ measured parallel to the line $\frac{x - 2}{2} = \frac{y + 2}{3} = \frac{z - 6}{-6}$ is (in $\text{ units}$)

Let the equation of the plane,that passes through the point $(1,4,-3)$ and contains the line of intersection of the planes $3x-2y+4z-7=0$ and $x+5y-2z+9=0$,be $\alpha x+\beta y+\gamma z+3=0$. Then $\alpha+\beta+\gamma$ is equal to:

If the line $\frac{x - 2}{3} = \frac{y + 1}{2} = \frac{z - 1}{-1}$ intersects the plane $2x + 3y - z + 13 = 0$ at a point $P$ and the plane $3x + y + 4z = 16$ at a point $Q$,then $PQ$ is equal to

If the line $\frac{x - 4}{1} = \frac{y - 2}{1} = \frac{z - k}{2}$ lies in the plane $2x - 4y + z = 7$,then $k = \dots$