If the sum of squares of distances of a point from the planes x + y...

Let distances of a point $P \left(x, y, z\right)$ from the planes $x + y + z = 0, x - z = 0$ and $x - 2y + z = 0$ are $\dfrac{x + y + z}{3}, \dfrac{x - z}{2}$ and $\dfrac{x - 2y + z}{6}$ respectively, then the sum of the squares of distances, is as

⇒ \left(\dfrac{x + y + z}{3}\right)2 + \left(\dfrac{x - z}{2}\right)2 + \left(\dfrac{x - 2y + z}{6}\right)2 = p^{2}$</p><p>$⇒ 2\left(x + y + z\right)^{2} + 3\left(x - z\right)^{2} + \left(x - 2y + z\right)^{2} = 6p^{2}$</p><p>$⇒ 2x^{2} + 2y^{2} + 2z^{2} + 4xy + 4yz + 4zx + 3x^{2} + 3z^{2} - 6xz + x^{2} + 4y^{2} + z^{2} - 4xy - 4yz + 2xz = 6p^{2}$</p><p>$⇒ 6x^{2} + 6y + 6z^{2} = 6p^{2}$</p><p>$⇒ x^{2} + y^{2} + z^{2} = p^{2}$</p><p>Hence, option (d) is correct.</p>$