The angle between the line $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z + 3}{-2}$ and the plane $x + y + 4 = 0$ is ......... $^o$.

The line of intersection of the planes $\vec r \cdot (3\hat i - \hat j + \hat k) = 1$ and $\vec r \cdot (\hat i + 4\hat j - 2\hat k) = 2$ is:

The line given by the equations $x-2y+4z+4=0$ and $x+y+z-8=0$ intersects the plane $x-y+2z+1=0$ at the point:

$A$ plane $E$ is perpendicular to the two planes $2x - 2y + z = 0$ and $x - y + 2z = 4$,and passes through the point $P(1, -1, 1)$. If the distance of the plane $E$ from the point $Q(a, a, 2)$ is $3\sqrt{2}$,then $(PQ)^2$ is equal to

The distance of the point $A(3, -4, 5)$ from the plane $2x + 5y - 6z = 16$ measured along the line $\frac{x}{2} = \frac{y}{1} = \frac{z}{-2}$ is

If a line $L$ is common to the planes $x-y+z+2=0$ and $2x+y-2z+5=0$,then the direction cosines of the line $L$ are