Find the equation of the plane that contains the point `(1, 1, 2)`and is perpendicular to each of the planes `2x + 3y 2z = 5`and `x + 2y 3z = 8`.

Explanation

The given equation of a plane passing through (1,-1,2) is a(x-1)+b(y+1)+c(z- 2) =0

A/Q;- It is perpendicular to the planes 2 x+3 y-2 z=5 and x+2 y-3 z=8.

Therefore, 2 a+3 b-2 c=0 and, a+2 b-3 c=0

On Solving these two equation by cross-multiplication, we get `frac{{a}}{-5}=frac{{b}}{4}=frac{{c}}{1}`

Substituting a=-5 c, b=4 c and c=1 in (i),

we get -5 x+4 y+z=-7 as the equation of the required plane.