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

We know that the equation of a plane passing through (1,−1,2) is `a(x−1)+b(y+1)+c(z−2)=0`

As It is perpendicular to the planes `2x+3y−2z=5` and `x+2y−3z=8`.

So,` 2a+3b−2c=0` and, `a+2b−3c=0`
Solving above two equation by cross-multiplication, we get `a/(-5)​=b/4​=c/1`
​
Substituting a=−5c,b=4c and c=1 in (i),

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