The distance of the point $(-1, -5, -10)$ from the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 5$ is:

If the equation of a plane $P,$ passing through the intersection of the planes $x+4y-z+7=0$ and $3x+y+5z=8$ is $ax+by+6z=15$ for some $a, b \in R,$ then the distance of the point $(3,2,-1)$ from the plane $P$ is

If the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{4}$ meets the plane $x + 2y + 3z = 15$ at a point $P$,then the distance of $P$ from the origin is

Let $L_1$ be the line of intersection of the planes given by the equations $2x+3y+z=4$ and $x+2y+z=5$. Let $L_2$ be the line passing through the point $P(2,-1,3)$ and parallel to $L_1$. Let $M$ denote the plane given by the equation $2x+y-2z=6$. Suppose that the line $L_2$ meets the plane $M$ at the point $Q$. Let $R$ be the foot of the perpendicular drawn from $P$ to the plane $M$. Then which of the following statements is (are) True?
$(A)$ The length of the line segment $PQ$ is $9\sqrt{3}$
$(B)$ The length of the line segment $QR$ is $15$
$(C)$ The area of $\triangle PQR$ is $\frac{3}{2}\sqrt{234}$
$(D)$ The acute angle between the line segments $PQ$ and $PR$ is $\cos^{-1}\left(\frac{1}{2\sqrt{3}}\right)$

The foot of the perpendicular drawn from $A(1, 2, 2)$ onto the plane $x+2y+2z-5=0$ is $B(\alpha, \beta, \gamma)$. If $\pi(x, y, z) \equiv x+2y+2z+5=0$ is a plane, then $-\pi(A) : \pi(B) =$ ?

The distance of the point having position vector $\hat{i}-2 \hat{j}-6 \hat{k}$ from the straight line passing through the point $(2, -3, -4)$ and parallel to the vector $6 \hat{i}+3 \hat{j}-4 \hat{k}$ is units.