Distance Of A Point From A Plane

The tool of choice is the point-to-plane distance formula: for a plane ax + by + cz + d = 0 and a point (x_1, y_1, z_1), the distance is |ax_1 + by_1 + cz_1 + d| / \sqrt{a^2 + b^2 + c^2}. This formula is a recurring JEE Advanced device built on projecting the displacement onto the unit normal. Substituting a = 2, b = -2, c = 1, d = 3 and the point (1, 2, -1): numerator = |2(1) - 2(2) + 1(-1) + 3| = |2 - 4 - 1 + 3| = |0| = 0. The denominator is \sqrt{4 + 4 + 1} = 3, so the distance is 0/3 = 0. A zero numerator means the point already satisfies the plane equation, so the point (1, 2, -1) lies on the plane and its perpendicular distance is 0. Option 2/3 misreads the constant term, option 1 drops a contribution, and option 2 mishandles the normal magnitude. This rests on the orthogonal-projection distance theorem. Plausibility check: substituting the point directly into 2x - 2y + z + 3 gives exactly 0, confirming the point is on the plane and the distance must vanish.