A variable plane passes through a fixed point | vedclass

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(B) Let the equation of the variable plane be $a(x - 1) + b(y - 2) + c(z - 3) = 0$,where $(a, b, c)$ are the direction ratios of the normal to the pla

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a sphere

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(B) Let the equation of the variable plane be $a(x - 1) + b(y - 2) + c(z - 3) = 0$,where $(a, b, c)$ are the direction ratios of the normal to the plane.Let $P(x_0, y_0, z_0)$ be the foot of the perpendicular from the origin $(0, 0, 0)$ to the plane.The vector $\vec{OP} = (x_0, y_0, z_0)$ is perpendicular to the plane,so it is parallel to the normal vector $(a, b, c)$.Thus,we can write $(a, b, c) = k(x_0, y_0, z_0)$ for some constant $k$.Since $P$ lies on the plane,it satisfies the equation: $x_0(x_0 - 1) + y_0(y_0 - 2) + z_0(z_0 - 3) = 0$.This simplifies to $x_0^2 - x_0 + y_0^2 - 2y_0 + z_0^2 - 3z_0 = 0$.Rearranging the terms,we get $(x_0^2 - x_0 + \frac{1}{4}) + (y_0^2 - 2y_0 + 1) + (z_0^2 - 3z_0 + \frac{9}{4}) = \frac{1}{4} + 1 + \frac{9}{4} = \frac{14}{4} = 3.5$.This is the equation of a sphere with center $(\frac{1}{2}, 1, \frac{3}{2})$ and radius $\sqrt{3.5}$.

$A$ variable plane passes through a fixed point $(1, 2, 3)$. The foot of the perpendicular from the origin to the plane lies on:

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