Let $L$ be the line of intersection of the planes $2x+3y+z=1$ and $x+3y+2z=1$. If $L$ makes an angle $\alpha$ with the positive $x$-axis,then $\cos \alpha$ equals:

The angle between the plane $x + y + z = 5$ and the line of intersection of the planes $3x + 4y + z - 1 = 0$ and $5x + 8y + 2z + 14 = 0$ is

The image of the line $\frac{x-1}{3}=\frac{y-3}{1}=\frac{z-4}{-5}$ in the plane $2x-y+z+3=0$ is the line

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

Let $\pi_1$ be the plane determined by the vectors $\bar{i}+\bar{j}$ and $\bar{i}+\bar{k}$, and $\pi_2$ be the plane determined by the vectors $\bar{j}-\bar{k}$ and $\bar{k}-\bar{i}$. Let $\bar{a}$ be a non-zero vector parallel to the line of intersection of the planes $\pi_1$ and $\pi_2$. If $\bar{b}=\bar{i}+\bar{j}-\bar{k}$, then the angle between the vectors $\bar{a}$ and $\bar{b}$ is:

Find the distance of the point $(2, 1, 0)$ from the plane $2x + y + 2z + 5 = 0$.