Let the foot of the perpendicular from the point $(1, 2, 4)$ on the line $\frac{x+2}{4} = \frac{y-1}{2} = \frac{z+1}{3}$ be $P$. Then the distance of $P$ from the plane $3x + 4y + 12z + 23 = 0$ is:

If the equation of the plane passing through the mirror image of a point $(2,3,1)$ with respect to the line $\frac{x+1}{2}=\frac{y-3}{1}=\frac{z+2}{-1}$ and containing the line $\frac{x-2}{3}=\frac{1-y}{2}=\frac{z+1}{1}$ is $\alpha x+\beta y+\gamma z=24$,then $\alpha+\beta+\gamma$ is equal to ..... .

The equation of the plane passing through the point of intersection of the planes $2x-y+z-3=0$ and $4x-3y+5z+9=0$ and parallel to the line $\frac{x+1}{2}=\frac{y+3}{4}=\frac{z-3}{5}$ is $\alpha x+\beta y+\gamma z+d=0$. Then $\alpha+\beta+\gamma+d=$

The equation of the plane containing the line $2x - 5y + z = 3; x + y + 4z = 5$ and parallel to the plane $x + 3y + 6z = 1$ is:

If $\bar{r} \cdot(2 \bar{i}+3 \bar{j}+4 \bar{k})=5$ and $\bar{r} \cdot(\bar{i}+\bar{j}-\bar{k})=7$ are two planes and $(16, -9, 0)$ is a point common to both the planes, then the vector equation of the line of intersection of the planes is $\bar{r}=$

The angle between the line with the direction ratios $(2, 5, 1)$ and the plane $8x + 2y - z = 4$ is