Question

Let the plane P contain the line 2x+y-z-3=0=5x-3y+4z+9 and be parallel to the line $\frac{x+2}{2}=\frac{3-y}{-4}=\frac{z-7}{5}$. Then the distance of the point
A(8, -1, -19) from the plane P measured parallel to the line $\frac{x}{-3}=\frac{y-5}{4}=\frac{2-z}{-12}$ is equal to ___________.

Answer 1

The answer is: 26