The equation of the plane which makes with coordinate axes, a triangle witih centroid `(alpha,beta,gamma)` is given by

Options

A. `alphax+betay+gammaz=1`
B. `alphax+betay+gammaz=3`
C. `x/alpha+y/beta=z/gamma=1`
D. `x/alpha+y/beta+z/gamma=3`

Answer

d

Explanation

Let the equation of the plane be `x/a+y/b+z/c=1`.
Then, it makes the axes at A(a,0,0), B(o,b,0) and C(0,0,c).
`therefore` centroid of `triangleABC` is `G((a+0+0)/3, (0+b+0)/3, (0+0+c)/3)`, i.e. `G(a/b,b/3,c/3)`.
`therefore (a/3=alpha, b/3=beta "and" c/3=lambda) rArr a=3alpha, b=3beta "and" c=3gamma`.
`therefore` the required equation of the plane is
`x/(3alpha)+y/(3beta)+z/(3gamma)=1 rArr x/alpha+y/beta+z/gama=3`.