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

The plane $2x - y + 3z + 5 = 0$ is rotated through $90^{\circ}$ about its line of intersection with the plane $x + y + z = 1$. The equation of the plane in its new position is:

Find the coordinates of the point of intersection of the line $\frac{x - 6}{-1} = \frac{y + 1}{0} = \frac{z + 3}{4}$ and the plane $x + y - z = 3$.

The vector equation of a plane passing through the line of intersection of the planes $\overline{r} \cdot(\overline{i}-2 \overline{k})=3$ and $\overline{r} \cdot(2 \overline{j}+\overline{k})=5$,and passing through the point $\overline{i}+2 \overline{j}+3 \overline{k}$,is:

The lines $\frac{x-0}{1}=\frac{y-2}{2}=\frac{z+3}{\lambda}$ and $\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{\lambda}$ are coplanar. If $p$ is the plane containing these lines,then which of the following points lies on the plane for all values of $\lambda$?

The equation of the plane passing through the origin and perpendicular to the line $x = 2y = 3z$ is