The equation of the plane passing through the point $(1,1,1)$ and through the line of intersection of $x+2y-z+1=0$ and $3x-y-4z+3=0$ is

$A$ plane $P$ contains the line $x+2y+3z+1=0=x-y-z-6$ and is perpendicular to the plane $-2x+y+z+8=0$. Then which of the following points lies on $P$?

The equation of the line of intersection of the planes $4x + 4y - 5z = 12$ and $8x + 12y - 13z = 32$ can be written as

Find the coordinates of the foot of the perpendicular drawn from the origin to the plane $5y + 8 = 0$.

Statement-$1$: The point $A(1, 0, 7)$ is the reflection of the point $B(1, 6, 3)$ in the line $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$.
Statement-$2$: The line $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$ is the perpendicular bisector of the line segment joining $A(1, 0, 7)$ and $B(1, 6, 3)$.

Let $P_1: 2x + y - z = 3$ and $P_2: x + 2y + z = 2$ be two planes. Then,which of the following statement$(s)$ is (are) $TRUE$?
$(A)$ The line of intersection of $P_1$ and $P_2$ has direction ratios $1, -1, 1$.
$(B)$ The line $\frac{3x - 4}{9} = \frac{1 - 3y}{9} = \frac{z}{3}$ is perpendicular to the line of intersection of $P_1$ and $P_2$.
$(C)$ The acute angle between $P_1$ and $P_2$ is $60^{\circ}$.
$(D)$ If $P_3$ is the plane passing through the point $(4, 2, -2)$ and perpendicular to the line of intersection of $P_1$ and $P_2$,then the distance of the point $(2, 1, 1)$ from the plane $P_3$ is $\frac{2}{\sqrt{3}}$.