If $n=2 \hat{i}-3 \hat{j}+4 \hat{k}$,$m=\hat{i}-\hat{j}$,and $l=2 \hat{i}-\hat{j}+\hat{k}$,then the Cartesian equation of the plane passing through the line of intersection of two planes $r \cdot n=1$ and $r \cdot m=-4$ and perpendicular to the plane $r \cdot l=-8$ is

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

The equation of the plane containing the line $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ and perpendicular to the plane containing the lines $\frac{x}{2}=\frac{y}{3}=\frac{z}{1}$ and $\frac{x}{3}=\frac{y}{2}=\frac{z}{1}$ is

If $P=(2,-3,4)$,$Q=(-1,-4,0)$,and $R=(2,1,0)$ are three points,and $S$ is the foot of the perpendicular drawn from $R$ to the line $PQ$,then the $X$-coordinate of $S$ is:

Let the line $L$ be the projection of the line $\frac{x-1}{2}=\frac{y-3}{1}=\frac{z-4}{2}$ in the plane $x-2y-z=3$. If $d$ is the distance of the point $(0,0,6)$ from $L$,then $d^2$ is equal to .... .

Let $P_1$ and $P_2$ be two planes given by $P_1: 10x + 15y + 12z - 60 = 0$ and $P_2: -2x + 5y + 4z - 20 = 0$. Which of the following straight lines can be an edge of some tetrahedron whose two faces lie on $P_1$ and $P_2$?
$(A) \frac{x-1}{0} = \frac{y-1}{0} = \frac{z-1}{5}$
$(B) \frac{x-6}{-5} = \frac{y}{2} = \frac{z}{3}$
$(C) \frac{x}{-2} = \frac{y-4}{5} = \frac{z}{4}$
$(D) \frac{x}{1} = \frac{y-4}{-2} = \frac{z}{3}$