Find the distance of the point $(-1,-5,-10)$ from the point of intersection of the line $\vec{r}=2 \hat{i}-\hat{j}+2 \hat{k}+\lambda(3 \hat{i}+4 \hat{j}+2 \hat{k})$ and the plane $\vec{r} \cdot(\hat{i}-\hat{j}+\hat{k})=5$.

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 $\cos^{-1}\left(\frac{1}{2\sqrt{3}}\right)$

Assertion $(A)$: The equation of the plane passing through the point $(4, 4, 4)$ and the intersection of the planes $x + y + z = 6$ and $2x + 3y + 4z = 0$ is $29x + 23y + 17z = 276$.
Reason $(R)$: The equation of the plane passing through the line of intersection of planes $P_1 = 0$ and $P_2 = 0$ is $P_1 + \lambda P_2 = 0, \lambda \in \mathbb{R}$.

The plane containing the point $(3,2,0)$ and the line $\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}$ is

The distance of the point $(-1, 9, -16)$ from the plane $2x + 3y - z = 5$ measured parallel to the line $\frac{x+4}{3} = \frac{2-y}{4} = \frac{z-3}{12}$ is $......$

$A$ line drawn from the point $A(1, 3, 2)$ parallel to the line $\frac{x}{2} = \frac{y}{4} = \frac{z}{1}$ intersects the plane $3x + y + 2z = 5$ at point $B$. Then the coordinates of point $B$ are: