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

A
5
B
$3 \sqrt{2}$
C
$\sqrt{21}$
D
$3 \sqrt{3}$

Solution:

Let P(x, y, z).Now under given condition, we get

$\begin{array}{l} {[\sqrt{x^{2} + y^{2}}]}^{2} + {[\sqrt{y^{2} + z^{2}}]}^{2} + {[\sqrt{z^{2} + x^{2}}]}^{2} = 36 \\ \Rightarrow x^{2} + y^{2} + z^{2} = 18 \end{array}$

Then distance from origin to the point (x, y, z) is

$\sqrt{x^{2} + y^{2} + z^{2}} = \sqrt{18} = 3 \sqrt{2}$

Solution:

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