If $\theta$ is the angle between the line $\frac{x + 1}{3} = \frac{y - 2}{2} = \frac{z - 2}{4}$ and the plane $2x + y - 3z + 4 = 0$,then $64 \csc^2 \theta$ is equal to

Let $\vec{r} \cdot (3\hat{i} - 2\hat{j} + 7\hat{k}) = 32$ be the equation of a plane. If the line with direction ratios $(5, b, 3)$ is parallel to the plane, then the value of $b$ is:

The coordinates of the point where the line joining $(1, 1, 1)$ and $(2, 2, 2)$ intersects the plane $x + y + z = 9$ are:

The equation of the plane containing the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$ and the point $(0,7,-7)$ is

The position vector of the point of intersection of the line $\bar{r}=(2 \hat{i}+\hat{j}-4 \hat{k})+\lambda(\hat{i}-2 \hat{j}+2 \hat{k})$ and the $XOY$-plane is:

$A$ line with positive direction cosines passes through the point $P(2,1,2)$ and makes equal angles with the coordinate axes. The line meets the plane $2x+y+z=9$ at point $Q$. The length of the line segment $PQ$ equals $\qquad$ units.