The distance of the point (1, 2, 4) from the plane passing through the point (1, 2, 2) and perpendicular to the planes x - y + 2z = 3 and 2x - y + z + 12 = 0 is

Let the equation of the plane beax+by+cz=dthen point(1,2,2)lies on the planeSo,a+2b+2c=d...eqn 1Since the plane is perpendicular to the given two planes, so dot product of the direction ratios of the perpendicular of the plane with the direction ratio of the perpendicular of the given two plane must be zero.So,a−b+2c=0and ...eqn 22a𕒶b+c=0eqn 3Doing eqn 1−eqn 2, we get3b=d,Similarly,On solving above 3 eqn we getc=0anda=bandd=3aPut the value ofb,canddin terms ofaand get the equation of the plane asx+y=3Henced=|1𕒶𕒷|√2=2√2