The angle between the line $\frac{x+1}{2}=\frac{y-2}{1}=\frac{z-3}{-2}$ and the plane $x-2y-\lambda z=3$ is $\cos^{-1}\left(\frac{2\sqrt{2}}{3}\right)$. Then the value of $\lambda$ is:

At what point does the line joining the points $(2, -3, 1)$ and $(3, -4, -5)$ intersect the plane $2x + y + z = 7$?

Let $Q$ and $R$ be the feet of perpendiculars from the point $P(a, a, a)$ on the lines $x=y, z=1$ and $x=-y, z=-1$ respectively. If $\angle QPR$ is a right angle,then $12a^2$ is equal to

Find the equation of the line passing through $(1, 1, 1)$ and perpendicular to the plane $2x + 3y - z - 5 = 0$.

$A$ straight line joining the points $(1, 1, 1)$ and $(0, 0, 0)$ intersects the plane $2x + 2y + z = 10$ at

Let $P$ be the plane containing the straight line $\frac{x-3}{9}=\frac{y+4}{-1}=\frac{z-7}{-5}$ and perpendicular to the plane containing the straight lines $\frac{x}{2}=\frac{y}{3}=\frac{z}{5}$ and $\frac{x}{3}=\frac{y}{7}=\frac{z}{8}$. If $d$ is the distance of $P$ from the point $(2,-5,11)$,then $d^{2}$ is equal to.