A point lies in the yz-plane,and the sum of i | vedclass

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(B) Let the coordinates of the point be $P(x, y, z)$.Since the point lies in the $yz$-plane,its $x$-coordinate must be $0$. So,$P = (0, y, z)$.The sum

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$(0, 2, 1)$

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(B) Let the coordinates of the point be $P(x, y, z)$.Since the point lies in the $yz$-plane,its $x$-coordinate must be $0$. So,$P = (0, y, z)$.The sum of the coordinates is $y + z = 3$ (Equation $1$).The distance of point $(0, y, z)$ from the $xz$-plane is $|y|$.The distance of point $(0, y, z)$ from the $xy$-plane is $|z|$.According to the problem,the distance from the $xz$-plane is twice the distance from the $xy$-plane,so $|y| = 2|z|$.Case $1$: $y = 2z$.Substituting into Equation $1$: $2z + z = 3 \implies 3z = 3 \implies z = 1$.Then $y = 2(1) = 2$.The point is $(0, 2, 1)$.Case $2$: $y = -2z$.Substituting into Equation $1$: $-2z + z = 3 \implies -z = 3 \implies z = -3$.Then $y = -2(-3) = 6$.The point is $(0, 6, -3)$.Comparing with the given options,the point $(0, 2, 1)$ is present.

$A$ point lies in the $yz$-plane,and the sum of its coordinates is $3$. If the distance of the point from the $xz$-plane is twice the distance of the point from the $xy$-plane,find its coordinates.

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