The value of $k$,such that the line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}$ lies on the plane $2x-4y+z=7$,is

Let the equation of the plane passing through the line $x-2y-z-5=0=x+y+3z-5$ and parallel to the line $x+y+2z-7=0=2x+3y+z-2$ be $ax+by+cz=65$. Then the distance of the point $(a, b, c)$ from the plane $2x+2y-z+16=0$ is $..........$.

The equation of the plane passing through the point $\hat{i}+2 \hat{j}-\hat{k}$ and perpendicular to the line of intersection of the planes $r \cdot(3 \hat{i}-\hat{j}+\hat{k})=1$ and $r \cdot(\hat{i}+4 \hat{j}-2 \hat{k})=2$ is:

If the image of the point $P(1, -2, 3)$ in the plane $2x + 3y - 4z + 22 = 0$ measured parallel to the line $\frac{x}{1} = \frac{y}{4} = \frac{z}{5}$ is $Q$,then $PQ$ is equal to:

The coordinates of the point where the line joining the points $(2, -3, 1)$ and $(3, -4, -5)$ intersects the plane $2x + y + z = 7$ are:

The coordinates of the point,where the line through $A(3, 4, 1)$ and $B(5, 1, 6)$ crosses the $XZ$-plane,are