The equation of the plane passing through the point (1, 1, 1) and perpendicular to the planes 2x + y - 2z = 5 and 3x - 6y -2z = 7, is

$14x + 2y +15z = 31$

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

It is perpendicular to the plane $2x + y - 2z = 5 $ and $ 3x - 6y -2z = 7.$

$∴ 2a + b - 2c = 0 $ and, $3a -6b -2c =0 $

$⇒ \frac{a}{-14}=\frac{b}{-2}=\frac{c}{-15}$

Substituting these values of a, b, c in (i), we obtain $14x + 2y + 15z = 31 $ as the equation of the required plane.