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 plan

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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 $Q(x_1, y_1, z_1)$ be the foot of the perpendicular from the origin $O(0, 0, 0)$ to the plane.Since $OQ$ is perpendicular to the plane, the direction ratios of $OQ$ are $(x_1, y_1, z_1)$, which must be proportional to $(a, b, c)$.Thus, $a = kx_1, b = ky_1, c = kz_1$ for some constant $k$.Substituting these into the plane equation: $x_1(x_1 - 1) + y_1(y_1 - 2) + z_1(z_1 - 3) = 0$.This simplifies to $x_1^2 + y_1^2 + z_1^2 - x_1 - 2y_1 - 3z_1 = 0$.This is the equation of a sphere with diameter $OP$, where $O$ is the origin and $P$ is the fixed point $(1, 2, 3)$.

$A$ variable plane passes through a fixed point $P(1, 2, 3)$. The foot of the perpendicular from the origin $O(0, 0, 0)$ to the plane lies on:

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