If d_1, d_2, d_3 are the distances of the poi | vedclass

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(B) The distance of a point $(x, y, z)$ from the $X, Y, Z$-axes are given by $d_1 = \sqrt{y^2 + z^2}$,$d_2 = \sqrt{x^2 + z^2}$,and $d_3 = \sqrt{x^2 +

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$2 d_1^2$

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(B) The distance of a point $(x, y, z)$ from the $X, Y, Z$-axes are given by $d_1 = \sqrt{y^2 + z^2}$,$d_2 = \sqrt{x^2 + z^2}$,and $d_3 = \sqrt{x^2 + y^2}$.For the point $(1, 2, 3)$:$d_1 = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \implies d_1^2 = 13$.$d_2 = \sqrt{1^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10} \implies d_2^2 = 10$.$d_3 = \sqrt{1^2 + 2^2} = \sqrt{1 + 4} = \sqrt{5} \implies d_3^2 = 5$.Now,calculate $2 d_2^2 + d_3^2 + 1$:$2(10) + 5 + 1 = 20 + 5 + 1 = 26$.Since $d_1^2 = 13$,we have $26 = 2 	imes 13 = 2 d_1^2$.

If $d_1, d_2, d_3$ are the distances of the point $(1, 2, 3)$ from the $X, Y, Z$-coordinate axes respectively,then $2 d_2^2 + d_3^2 + 1 =$

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