A variable plane which remains at q constant distance `3p` from the origin cut the coordinate axes at `A ,\ B ,\ Cdot` Show that the locus o the centroid of triangle `A B C` is `x^(-2)+y^(-2)+z^(-2)=p^(-2)dot`

Explanation

Let the equation of variable plane be
`x/a+y/b+z/c = 1 "……"(1)`
Which meets the axes at `A(a,0,0), B(0,b,0)` and `C(0,0,c)` respectively.
Let the centroid of `DeltaABC` is `(alpha,beta,gamma)`
`alpha = (a+0+0)/(3),beta = (0+b+0)/(3), gamma = (0+0+c)/(3)`
`rArr alpha = 3alpha , b = 3beta, c = 3gamma "......"(2)`
Now the length of perpendicular from `(0,0,0)` to plane `(1) = 3p`
`rArr |(0+0+0-1)/(sqrt(1/(a^(2))+(1)/(b^(2))+(1)/(c^(2))))|=3p`
`rArr (1)/(a^(2))+(1)/(b^(2))+(1)/(c^(2))=(1)/(9p^(2))`
`rArr 1/(9alpha^(2)) + (1)/(9beta^(2)) + (1)/(9gamma^(2)) = (1)/(9p^(2))` [From eq. (2)]
`rArr alpha^(-2) + beta^(-2) + gamma^(-2) = p^(-2)`
`:.` Locus of `(alpha, beta, gamma)`
`x^(-2)+y^(-2)+z^(-2) = p^(-2)`.