A plane passes through (1, -2, 1) and is perpendicular to two planes 2x - 2y + z = 0 and x - y + 2z = 4. The distance of the plane from the point (1, 2, 2) is
Answer & explanation
Correct answer: option 4
The equation of a plane passing through (1, -2, 1) is
$a(x-1) + b(y+2)+c(z-1) = 0 $ .........(i)
It is perpendicular to 2x - 2y + z = 0 and x - y + 2z = 4.
$∴ 2a - 2b + c = 0 $ and $ a - b + 2c = 0 $
$ \frac{a}{-3}=\frac{b}{-3}=\frac{c}{0}$
Substituting the values of a, b, c in (i), we obtain
$ x + y + 1= 0 $
Distance of this plane from (1, 2, 2) is given by
$d = \frac{1+2+1}{\sqrt{1+1}}= 2\sqrt{2}$