Perpendiculars are drawn from points on the line $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z}{3}$ to the plane $x+y+z=3$. The feet of the perpendiculars lie on the line:

Find the distance of the point $-\hat{i} - 5\hat{j} - 10\hat{k}$ from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 5$.

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

$\vec{r} \cdot(\hat{i}-\hat{j}+\hat{k})=5$ and $\vec{r} \cdot(2 \hat{i}+\hat{j}-\hat{k})=3$ are two planes. $A$ plane $\pi$ passing through the line of intersection of these two planes, passes through the point $(0,1,2)$. If the equation of $\pi$ is $\vec{r} \cdot(a \hat{i}+b \hat{j}+c \hat{k})=m$, then $\frac{b c}{a^2}=$

If the points $(1, 1, \lambda)$ and $(-3, 0, 1)$ are equidistant from the plane $3x + 4y - 12z + 13 = 0$, then the values of $\lambda$ are

Let $\bar{A}$ be a vector parallel to the line of intersection of planes $P_1$ and $P_2$ passing through the origin. $P_1$ is parallel to the vectors $2 \hat{j}+3 \hat{k}$ and $4 \hat{j}-3 \hat{k}$,and $P_2$ is parallel to $\hat{j}-\hat{k}$ and $3 \hat{i}+3 \hat{j}$. Then the angle between $\bar{A}$ and $2 \hat{i}+\hat{j}-2 \hat{k}$ is