The locus of a first degree equation in $x, y, z$ is a

In three-dimensional space,the equation $3y + 4z = 0$ represents:

The plane passing through the point $(4, -1, 2)$ and parallel to the lines $\frac{x + 2}{3} = \frac{y - 2}{-1} = \frac{z + 1}{2}$ and $\frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 4}{3}$ also passes through the point

If a plane passing through the points $(2,3,0), (0,-5,2)$ and $(-2,0,3)$ meets the $X, Y, Z$-axes in $A, B, C$ respectively,then $A=$

The equation of the plane,passing through the point $(1, 1, 1)$ and perpendicular to the planes $2x + y - 2z = 5$ and $3x - 6y - 2z = 7$,is

If the angle between the planes $x-2y+3z-5=0$ and $x+\alpha y+2z+7=0$ is $\cos^{-1}\left(\frac{1}{14}\right)$,then the difference between the values of $\alpha$ is