Find the angle between the line $\vec{r} = (2\hat{i} - \hat{j} + \hat{k}) + \lambda(-\hat{i} + \hat{j} + \hat{k})$ and the plane $\vec{r} \cdot (3\hat{i} + 2\hat{j} - \hat{k}) = 4$.

The equation of a plane through the line of intersection of the planes $x + 2y = 3$ and $y - 2z + 1 = 0$,and perpendicular to the first plane $x + 2y = 3$ is:

Let $\theta$ be the angle between the planes $P_1=\vec{r} \cdot(\hat{i}+\hat{j}+2\hat{k})=9$ and $P_2=\vec{r} \cdot(2\hat{i}-\hat{j}+\hat{k})=15$. Let $L$ be the line that meets $P_2$ at the point $(4,-2,5)$ and makes an angle $\theta$ with the normal of $P_2$. If $\alpha$ is the angle between $L$ and $P_2$,then $(\tan^2 \theta)(\cot^2 \alpha)$ is equal to $...........$.

Find the equation of the plane passing through the point $(3, 2, 0)$ and the line $\frac{x - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4}$.

What is the equation of the line passing through $(1, 2, 3)$ and perpendicular to the plane $3x + 4y - 5z = 6$?

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