The angle between the line $\frac{x + 1}{3} = \frac{y - 1}{4} = \frac{z - 2}{2}$ and the plane $2x - 3y + z + 4 = 0$ is:

If the angle between the line $x = \frac{y-1}{2} = \frac{z-3}{\lambda}$ and the plane $x + 2y + 3z = 4$ is $\cos^{-1} \sqrt{\frac{5}{14}}$,then the value of $\lambda$ is

The foot of the perpendicular drawn from a point $A(1,1,1)$ onto a plane $\pi$ is $P(-3,3,5)$. If the equation of the plane parallel to the plane $\pi$ and passing through the midpoint of $AP$ is $ax-y+cz+d=0$,then $a+c-d=$

If the angle between the line $x = \frac{y - 1}{2} = \frac{z - 3}{\lambda}$ and the plane $x + 2y + 3z = 4$ is $\cos^{-1}\left(\sqrt{\frac{5}{14}}\right)$,then $\lambda = \dots$

For $a, b \in \mathbb{Z}$ and $|a - b| \leq 10$,let the angle between the plane $P: ax + y - z = b$ and the line $l: x - 1 = \frac{-y}{1} = z + 1$ be $\cos^{-1}\left(\frac{1}{3}\right)$. If the distance of the point $(6, -6, 4)$ from the plane $P$ is $3\sqrt{6}$,then $a^4 + b^2$ is equal to

The position vector of the point in which the line joining the points $i - 2j + k$ and $3k - 2j$ cuts the plane through the origin and the points $4j$ and $2i + k$,is