Question

Find the equation of plan which passes through the point (5,-3,2) and is perpendicular to each of planes x-y+z=0 and x=y-z=6

Answer 1

The new plane must have a direction vectors of <1,-1,1> and <1,1,-1>
(I suspect that your meant x+y-z=6 for your second plane equation)

Therefore the normal of the new plane is the cross-product of those two.
That normal would be <0,1,1>
So your new plane equation is y + z = c
given: the point (5,-3,2) is on that plane, then
-3 + 2 = c
c = -1

new equation: y + z = -1

in vector notation, you could just have written
r = (5,-3,2) + s(1,-1,1) + t(1,1,-1)