The mirror image of the point $P(-1, 2, -4)$ in the plane $x - y - 2z + 1 = 0$ is

The direction ratios of the two lines $AB$ and $AC$ are $1, -1, -1$ and $2, -1, 1$. The direction ratios of the normal to the plane $ABC$ are

$A$ plane $\Pi$ passes through the points $A=(0,0,2)$, $B=(1,0,1)$, and $C=(3,1,1)$. If the plane $\Pi$ makes angles $\alpha$ and $\beta$ with the $XY$ and $XZ$-coordinate planes respectively, then $\sin^2 \alpha + \sin^2 \beta =$

The equation of the plane which passes through $(2, -3, 1)$ and is normal to the line joining the points $(3, 4, -1)$ and $(2, -1, 5)$ is given by:

Let $\pi_1$ be the plane passing through the point $2\hat{i}-\hat{j}+\hat{k}$ and perpendicular to the vector $a\hat{i}+2\hat{j}-3\hat{k}$, and $\pi_2$ be the plane passing through the point $\hat{i}+2\hat{j}-\hat{k}$ and perpendicular to the vector $\hat{i}-2\hat{j}+\hat{k}$. If $\theta$ is the angle between the planes $\pi_1$ and $\pi_2$ and $\cos \theta = -\sqrt{\frac{3}{7}}$, then the integral value of $a$ is:

Find the angle between the planes $2x - y + z = 6$ and $x + y + 2z = 3$.