The equation of the plane perpendicular to the line $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ and passing through the point $(2, 3, 4)$ is:

In space,the equation $by + cz + d = 0$ represents a plane perpendicular to the plane:

If $\overrightarrow{p} = 4\hat{i} - \hat{j} + \hat{k}$ is a point and $\overrightarrow{q} = 9\hat{i} - 2\hat{j} + 6\hat{k}$ is a normal vector,then the perpendicular distance of the origin from the plane passing through $\overrightarrow{p}$ and perpendicular to $\overrightarrow{q}$ is

The equation of the plane passing through $(1, 2, 3)$ and parallel to the plane $2x + 3y - 4z = 0$ is

$A$ plane $E_{1}$ makes intercepts $1, -3, 4$ on the coordinate axes. The equation of a plane parallel to plane $E_{1}$ and passing through $(2, 6, -8)$ is

In three-dimensional space,if $f(x, y, z) = xy + xz$,then the locus of all points which satisfy the equation $f(x, y, z) = 0$ is -