Let $P$ be the image of the point $(3,1,7)$ with respect to the plane $x-y+z=3$. Then the equation of the plane passing through $P$ and containing the straight line $\frac{x}{1}=\frac{y}{2}=\frac{z}{1}$ is

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 $P_2$?
$(A) \frac{x-1}{0} = \frac{y-1}{0} = \frac{z-1}{5}$
$(B) \frac{x-6}{-5} = \frac{y}{2} = \frac{z}{3}$
$(C) \frac{x}{-2} = \frac{y-4}{5} = \frac{z}{4}$
$(D) \frac{x}{1} = \frac{y-4}{-2} = \frac{z}{3}$

The equation of the plane containing the straight line $\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3}=\frac{y}{4}=\frac{z}{2}$ and $\frac{x}{4}=\frac{y}{2}=\frac{z}{3}$ is:

Find the equation of the plane which passes through the points $(0,1,2)$ and $(-1,0,3)$ and is perpendicular to the plane $2x+3y+z=5$.

Consider the planes $3x - 6y - 2z = 15$ and $2x + y - 2z = 5$.
$STATEMENT-1$ : The parametric equations of the line of intersection of the given planes are $x = 3 + 14t, y = 1 + 2t, z = 15t$ because
$STATEMENT-2$ : The vector $14\hat{i} + 2\hat{j} + 15\hat{k}$ is parallel to the line of intersection of the given planes.

The line $\frac{x + 3}{3} = \frac{y - 2}{-2} = \frac{z + 1}{1}$ and the plane $4x + 5y + 3z - 5 = 0$ intersect at a point