If the sum of the squares of the distance of a point from the three coordinate axes be 36, then its distance from the origin is

. Let P(x, y,z)be any point in the plane. According to the given condition, (distance from P to X-axis)2 + (distance from P to Y-axis)2 + (distance from P to Z-axis)2 = 36⇒    y2+z22+x2+z22+x2+y22=36⇒    y2+z2+x2+z2+x2+y2=36⇒    2x2+y2+z2=36⇒    x2+y2+z2=18----iThe distance from origin to the point (x, y, z) is=x2+y2+z2=18 [∵ from Eq. (i)] =32