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 $units$)

Let the image of the point $\left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right)$ in the plane $x-2y+z-2=0$ be $P$. If the distance of the point $Q(6, -2, \alpha)$,where $\alpha > 0$,from $P$ is $13$,then $\alpha$ is equal to $...........$.

If the plane $3x - 4y - kz = 7$ contains the line $\frac{1 - x}{-2} = \frac{y + 1}{3} = \frac{z}{4}$,find the value of $k$.

What is the distance of the point $(2, 3, 4)$ from the plane $3x - 6y + 2z + 11 = 0$?

If the equation of the plane passing through the point $(2,-3,4)$ and perpendicular to both the planes $2x-3y+5z=2$ and $x+y+2z=3$ is $x+py+qz=r$, then $r$ is equal to

If the lines $\frac{x - a + d}{\alpha - \delta} = \frac{y - a}{\alpha} = \frac{z - a - d}{\alpha + \delta}$ and $\frac{x - b + c}{\beta - \gamma} = \frac{y - b}{\beta} = \frac{z - b - c}{\beta + \gamma}$ are coplanar,then the equation of the plane containing them is .........