The equation of a plane containing the line of intersection of the planes $2x - y - 4 = 0$ and $y + 2z - 4 = 0$ and passing through the point $(1, 1, 0)$ is

The equation of the line passing through $(1, 2, 3)$ and parallel to the planes $x - y + 2z = 5$ and $3x + y + z = 6$ is:

If the angle $\theta$ between the line $\frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}$ and the plane $2x-y+\sqrt{\lambda}z+4=0$ is such that $\sin \theta=\frac{1}{3}$,then $\lambda+1=$

Equation of the plane which passes through the point of intersection of lines $\frac{x - 1}{3} = \frac{y - 2}{1} = \frac{z - 3}{2}$ and $\frac{x - 3}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ and has the largest distance from the origin is

The mirror image of $P(2, 4, -1)$ in the plane $x - y + 2z - 2 = 0$ is $Q(a, b, c)$. Then the value of $a + b + c$ is:

Consider the line $L$ given by the equation $\frac{x-3}{2}=\frac{y-1}{1}=\frac{z-2}{1}$. Let $Q$ be the mirror image of the point $P_0(2,3,-1)$ with respect to $L$. Let a plane $P$ be such that it passes through $Q$,and the line $L$ is perpendicular to $P$. Then which of the following points is on the plane $P$?