A variable plane which remains at a constant distance 3p from the origin cuts the coordinate axes at A, B, C. Show that the locus of the centroid of triangle ABC is 1/x² + 1/y² + 1/z² = 1/p²<p></p>

Let the equation of plane be xa + yb + zc = 1, where a, b, c are intercepts of plane on x, y, z axis respectively.The distance from origin to plane will be 1/√(1/a² + 1/b² + 1/c²) = 3p ⇒ 1/a² + 1/c² + 1/b² = 1/9p²The centroid of triangle ABC is (a/3, b/3, c/3) and let it be (x, y, z)So we have a = 3x, b = 3y, c = 3z ⇒ 1/9x² + 1/9y² + 1/9z² = 1/9p² ⇒ 1/x² + 1/y² + 1/z² = 1/p²Hence proved

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Question:

A variable plane which remains at a constant distance 3p from the origin cuts the coordinate axes at A, B, C. Show that the locus of the centroid of triangle ABC is 1/x² + 1/y² + 1/z² = 1/p²

Solution: