Question

If the centroid of the tetrahedron $OABC$, where $A,B,C$ are given by $(\alpha ,5,6),(1,\beta ,4),$ $(3,2,\gamma )$ respectively is $(1,-1,2)$, then value of ${\alpha }^2+{\beta }^2+{\gamma }^2$ equals

• A. ${\alpha }^2+{\beta }^2$
• B. ${\gamma }^2+{\beta }^2$
• C. ${\alpha }^2+{\gamma }^2$
• D. ${\alpha }^3+{\beta }^3$

Answer 1

The correct option is B ${\gamma }^2+{\beta }^2$

The centroid of tetrahedron having vertices $O(0,0,0),A(\alpha ,5,6),B(1,\beta ,4)$ and $D(3,2,\gamma )$ is

$=(\frac{0+\alpha +1+3}{4},\frac{0+5+\beta +2}{4},\frac{0+6+4+\gamma }{4})=(1,-1,2)$ (given)

$\Rightarrow (\frac{\alpha +4}{4},\frac{\beta +7}{4},\frac{\gamma +10}{4})=(1,-1,2)$

Equating respective coordinate, we get $\alpha =0,\beta =-11,\gamma =-2$

So clearly ${\alpha }^2+{\beta }^2+{\gamma }^2={\beta }^2+{\gamma }^2$, since $\alpha =0$.