If the centroid of triangle whose vertices are $(a, 1,3)$, $(-2, \mathrm{~b},-5)$ and $(4,7, \mathrm{c})$ be…
If the centroid of triangle whose vertices are $(a, 1,3)$, $(-2, \mathrm{~b},-5)$ and $(4,7, \mathrm{c})$ be the origin, then $\mathrm{a}^2+\mathrm{b}^2+\mathrm{c}^2=$
$68$
$64$
$72$
$54$
Solution
Centroid of triangle with vertices $\left(x_1, y_1, z_1\right),\left(x_2, y_2\right.$,$\left.z_2\right)$ and $\left(x_3, y_3, z_3\right)$ is
$\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$
For $(a, 1,3),(-2, b,-5)(4,7, c)$;
if centroid is $(0,0,0)$
$\Rightarrow 0=\frac{a-2+4}{3} \Rightarrow a=-2$
$\begin{aligned} & 0=\frac{1+b+7}{3} \Rightarrow b=-8 \\ & 0=\frac{3-5+c}{3} \Rightarrow c=2\end{aligned}$
$\begin{aligned} & a^2+b^2+c^2 \\ & =(-2)^2+(-8)^2+2^2 \\ & =4+64+4 \\ & =72\end{aligned}$