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Foot of the Perpendicular from a Point on a Plane

Let Pa,b,c be...

Question

Let $P (a, b, c)$ be any point on the plane $3 x + 2 y + z = 7.$ Then the least value of $2 (a^{2} + b^{2} + c^{2})$ is

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Solution

If $a^{2} + b^{2} + c^{2}$ is minimum,
then $P (a, b, c)$ is the foot of the perpendicular from the origin to the given plane.

$\Rightarrow (a^{2} + b^{2} + c^{2})_{min} =$ Square of perpendicular distance from the origin to the given plane.
$\Rightarrow (a^{2} + b^{2} + c^{2})_{min} = \frac{49}{14} = \frac{7}{2}$

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