The equation of the plane which makes with co-ordinate axes, a triangle with its centroid (α,β,γ)is

[c] Let us take a triangle ABC and their vertices A (a, 0, 0), B(0, b, 0) and C(0, 0, c)$Theref\text{or}etheequationofpla\ne is$xa+yb+zc=1                                                   (i)$Now,givencentroidof\Delta ABCis\left(\alpha ,\beta ,\gamma \right)Asweknow,centroidof\Delta ABCwithvertices\left(x1,y1,z1\right),\left(x2,y2,z2\right)\text{and}\left(x3,y3,z3\right)isgivenby\left(x1+x2+x33,y1+y2+y33,z1+z2+z33,\right)$∴ By using this formula, we have$a+0+03=\alpha \Rightarrow a=3\alpha ,;0+b+b3=\beta$⇒b=3β$And0+0+c3=\gamma \Rightarrow c=3\gamma$Now, put the values of a, b, c, in equation (i), which gives$x3\alpha +y3\beta +z3\gamma =1\therefore x\alpha +y\beta +z\gamma =3$