Distance Between Parallel Planes

The applicable result is the distance between parallel planes ax + by + cz = d_1 and ax + by + cz = d_2, given by |d_1 - d_2| / \sqrt{a^2 + b^2 + c^2}. This compact formula is a routine JEE Advanced fact, valid only after the normal coefficients are made identical, which they already are here. With a = 2, b = -1, c = 2, the normal magnitude is \sqrt{4 + 1 + 4} = 3. The constants are d_1 = 5 and d_2 = 11, so |d_1 - d_2| = 6, and the distance is 6/3 = 2. Option 3 mistakenly divides by 2 instead of 3. Option 6 forgets to divide by the normal magnitude entirely. Option 1 halves the constant difference incorrectly. This relies on the parallel-plane distance theorem, which is the difference of constant terms projected onto the unit normal. Plausibility check: the planes share the same normal direction so they are genuinely parallel, and the positive value 2 represents a consistent uniform gap between them everywhere in space.