The reflection of the point $(-1, 3, 4)$ with respect to the plane $x - 2y = 0$ is .....

The distance of the point $(3, 4, 5)$ from the point of intersection of the line $\frac{x-3}{1} = \frac{y-4}{2} = \frac{z-5}{2}$ and the plane $x+y+z=2$ is: (in $units$)

The plane $lx + my = 0$ is rotated by an angle $\alpha$ about its line of intersection with the plane $z = 0$. Find the equation of the plane in its new position.

Find the equation of the plane passing through the intersection of the planes $P_1$ and $P_2$ and parallel to the line $L$,where:
$P_1 : 3x + 2y + 5z + 1 = 0$
$P_2 : x + y + z + 2 = 0$
$L : \frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}$

The equation of the plane passing through the lines $\frac{x - 4}{1} = \frac{y - 3}{1} = \frac{z - 2}{2}$ and $\frac{x - 3}{1} = \frac{y - 2}{-4} = \frac{z}{5}$ is

Let the plane $P: \vec{r} \cdot \vec{a} = d$ contain the line of intersection of two planes $\vec{r} \cdot (\hat{i} + 3\hat{j} - \hat{k}) = 6$ and $\vec{r} \cdot (-6\hat{i} + 5\hat{j} - \hat{k}) = 7$. If the plane $P$ passes through the point $(2, 3, 1/2)$,then the value of $\frac{|13\vec{a}|^2}{d^2}$ is equal to