The equation of the plane passing through a point $A(2, -1, 3)$ and parallel to the vectors $\vec{a} = (3, 0, -1)$ and $\vec{b} = (-3, 2, 2)$ is:

The Cartesian equation of the plane passing through the point $(1, -2, 3)$ and perpendicular to the vector $-\hat{i} + 2\hat{j} - 3\hat{k}$ is:

Let two planes be $P_1 : 2x - y + z = 2$ and $P_2 : x + 2y - z = 3$. Based on the given information,the equation of the acute angle bisector of the planes $P_1$ and $P_2$ is...

Find the vector equation of a plane which is at a distance of $7$ units from the origin and normal to the vector $3 \hat{i} + 5 \hat{j} - 6 \hat{k}$.

$A$ vector $\vec{n}$ is inclined to the $x$-axis at $45^\circ$,to the $y$-axis at $60^\circ$,and at an acute angle to the $z$-axis. If $\vec{n}$ is a normal to a plane passing through the point $(\sqrt{2}, -1, 1)$,then the equation of the plane is:

If the plane $\frac{x}{2} - \frac{y}{3} - \frac{z}{5} = 1$ cuts the coordinate axes at points $A, B,$ and $C$ respectively,then the area of the triangle $ABC$ is: