Question

A variable plane which remains at a constant distance 3p from the origin, cuts the co-ordinate axes at A, B, C. Show that the locus of the centroid of the triangle ABC is $1 over straight x squared plus 1 over straight y squared plus 1 over straight z squared space equals 1 over straight p squared.$

Solution

Let O be the origin and OA = a, OB = b, OC = c
∴ equation of plane passing through A, B and C is
$straight x over straight a plus straight y over straight b plus straight z over straight c equals 1$
or $straight x over straight a plus straight y over straight b plus straight z over straight c minus 1 space equals space 0$
From the given condition, $fraction numerator open vertical bar 0 plus 0 plus 0 minus 1 close vertical bar over denominator square root of begin display style 1 over straight a squared end style plus begin display style 1 over straight b squared end style plus begin display style 1 over straight c squared end style end root end fraction space equals space 3 space straight p$
$rightwards double arrow space space space space space space space space space square root of 1 over straight a squared plus 1 over straight b squared plus 1 over straight c squared end root space equals space fraction numerator 1 over denominator 3 space straight p end fraction rightwards double arrow space space space space space space space 1 over straight a squared plus 1 over straight b squared plus 1 over straight c squared space equals fraction numerator 1 over denominator 9 space straight p squared end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis$

Now A, B, C are (a, 0, 0), (0, b, 0), (0, c, 0) respectively.
Let (x₁ , y₁ , z₁ ) be the centroid of ΔABC.
$therefore space space space straight x subscript 1 space equals space fraction numerator straight a plus 0 plus 0 over denominator 3 end fraction comma space space space straight y subscript 1 space equals space fraction numerator 0 plus straight b plus 0 over denominator 3 end fraction comma space straight z subscript 1 space equals space fraction numerator 0 plus 0 plus 0 over denominator 3 end fraction therefore space space space space straight a space equals space 3 space straight x subscript 1 comma space space space space straight b space equals space 3 space straight y subscript 1 comma space space space straight c space equals space 3 space straight z subscript 1$