The Return

The normal vector of both the planes is $i -3j+3k$ .

Assume any point on both the planes(A and B), calculate the distance between the two points, and mutiply it by $\cos\theta$ to get the answer, where $\theta$ is the angle between the normal vector and the $AB$ vector.

For simplicity's sake, let $A(8,0,0)$ and $B(1,0,0)$ .

$|AB|=7, BA=7i$

$\cos\theta=\frac {7-0+0}{√7\cdot \sqrt {1+9+9}}$ .

Thus, the distance between the two planes is $\frac {7}{√19}≈1.605$ .

Divide the second plane equation by 2 to get the same normal vector (for easy calculations). Ans is |8-1|/(length of normal vector)