A point moves such that the sum of the square | vedclass

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(C) Let the coordinates of the point be $P(x, y, z)$.The distance of point $P(x, y, z)$ from the $x$-axis is $\sqrt{y^2 + z^2}$. The square of this di

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$3\sqrt{2}$

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(C) Let the coordinates of the point be $P(x, y, z)$.The distance of point $P(x, y, z)$ from the $x$-axis is $\sqrt{y^2 + z^2}$. The square of this distance is $y^2 + z^2$.The distance of point $P(x, y, z)$ from the $y$-axis is $\sqrt{x^2 + z^2}$. The square of this distance is $x^2 + z^2$.The distance of point $P(x, y, z)$ from the $z$-axis is $\sqrt{x^2 + y^2}$. The square of this distance is $x^2 + y^2$.According to the problem,the sum of the squares of these distances is $36$:$(y^2 + z^2) + (x^2 + z^2) + (x^2 + y^2) = 36$Simplifying the equation:$2x^2 + 2y^2 + 2z^2 = 36$$x^2 + y^2 + z^2 = 18$The distance of the point $P(x, y, z)$ from the origin $(0, 0, 0)$ is given by $d = \sqrt{x^2 + y^2 + z^2}$.Substituting the value:$d = \sqrt{18} = \sqrt{9 	imes 2} = 3\sqrt{2}$.

$A$ point moves such that the sum of the squares of its distances from the coordinate axes is $36$. Find the distance of this point from the origin.

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