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(B) Let the origin be $O(0, 0, 0)$ and the foot of the perpendicular be $P(1, 2, 3)$.Since $OP$ is perpendicular to the plane,the direction ratios of

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$(7, 2, 1)$

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(B) Let the origin be $O(0, 0, 0)$ and the foot of the perpendicular be $P(1, 2, 3)$.Since $OP$ is perpendicular to the plane,the direction ratios of the normal to the plane are the same as the direction ratios of the line segment $OP$.The direction ratios of $OP$ are $\langle 1-0, 2-0, 3-0 \rangle = \langle 1, 2, 3 \rangle$.Thus,the normal vector to the plane is $\vec{n} = \hat{i} + 2\hat{j} + 3\hat{k}$.The equation of a plane passing through a point $(x_1, y_1, z_1)$ with normal vector $\langle a, b, c \rangle$ is given by $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$.Substituting the point $P(1, 2, 3)$ and the normal vector $\langle 1, 2, 3 \rangle$:$1(x - 1) + 2(y - 2) + 3(z - 3) = 0$$x - 1 + 2y - 4 + 3z - 9 = 0$$x + 2y + 3z - 14 = 0$.Now,we check which of the given options satisfies this equation:For option $(B) (7, 2, 1)$:$7 + 2(2) + 3(1) - 14 = 7 + 4 + 3 - 14 = 14 - 14 = 0$.Since the point $(7, 2, 1)$ satisfies the equation of the plane,it lies on the plane.

If the foot of the perpendicular drawn from the origin to a plane is $(1, 2, 3)$,then a point on that plane is

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