The equation of a plane containing the line $\frac{x + 1}{-3} = \frac{y - 3}{2} = \frac{z + 2}{1}$ and the point $(0, 7, -7)$ 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 acute angle between the line $r = (-\hat{i} + 3\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$ and the plane $r \cdot (10\hat{i} + 2\hat{j} - 11\hat{k}) = 3$ is:

If the distance between the plane $Ax - 2y + z = d$ and the plane containing the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\frac{x-2}{3} = \frac{y-3}{4} = \frac{z-4}{5}$ is $\sqrt{6}$,then $|d|$ is

The foot of the perpendicular drawn from the point $(4,2,3)$ to the line joining the points $(1,-2,3)$ and $(1,1,0)$ lies on the plane

From a point $P(\lambda, \lambda, \lambda)$,perpendiculars $PQ$ and $PR$ are drawn respectively on the lines $y=x, z=1$ and $y=-x, z=-1$. If $P$ is such that $\angle QPR$ is a right angle,then the possible value$(s)$ of $\lambda$ is(are)