Question

if the distance between the plane Ax-2y+z=d and the plane containing the lines
$\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ and $\frac{x-2}{3} = \frac{y-3}{4} = \frac{z-4}{5}$ is $\sqrt{6}$ then $|d|=$

Answer 1

i have calculated A=1
rest u can do it
try it