$A$ line with positive direction cosines passes through the point $P(2,-1,2)$ and makes equal angles with the coordinate axes. The line meets the plane $2x+y+z=9$ at point $Q$. The length of the line segment $PQ$ is equal to:

The angle between the line $r = (2i - j + k) + \lambda (-i + j + k)$ and the plane $r \cdot (3i + 2j - k) = 4$ is:

The equation of the plane passing through the line $\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-3}{4}$ and perpendicular to the plane $x+2y+z=12$ is given by $ax+by+cz+4=0$. Then:

Let the foot of the perpendicular from the point $A(4, 3, 1)$ on the plane $P: x - y + 2z + 3 = 0$ be $N$. If $B(5, \alpha, \beta)$,where $\alpha, \beta \in \mathbb{Z}$,is a point on the plane $P$ such that the area of the triangle $ABN$ is $3\sqrt{2}$,then $\alpha^2 + \beta^2 + \alpha\beta$ 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 $\alpha+\beta+\gamma+d=$

Find the angle between the line $\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}$ and the plane $10x+2y-11z=3$.