The angle between the line $\frac{x - 2}{a} = \frac{y - 2}{b} = \frac{z - 2}{c}$ and the plane $ax + by + cz + 6 = 0$ is ......... $^o$

The distance of the line $\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}$ from the point $(1,4,0)$ along the line $\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}$ is:

The ratio in which the line joining $(2, -4, 3)$ and $(-4, 5, -6)$ is divided by the plane $3x + 2y + z - 4 = 0$ is

The position vectors of the points $A$ and $B$ are respectively $\hat{i}+2 \hat{j}$ and $2 \hat{i}+\hat{j}+\hat{k}$. If the points $P$ and $Q$ are respectively the orthogonal projections of $A$ and $B$ on the plane $x+y+z=3$,then $P Q=$

If the line $\vec{r} = (\hat{i} - 2\hat{j} - \hat{k}) + \lambda (2\hat{i} + \hat{j} + 2\hat{k})$ is parallel to the plane $\vec{r} \cdot (3\hat{i} - 2\hat{j} - m\hat{k}) = 14$,then the value of $m$ is:

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