If the origin is the centroid of triangle

Solution:

Since, centroid of a triangle is found by

$\left(\frac{x_{2}+x_{1}+x_{1}}{3}, \frac{y_{2}+y_{1}+y_{2}}{3}, \frac{z_{2}+z_{1}+z_{3}}{3}\right)$

The points are $A(a, 1,3)$ and $B(-2, b,-5)$, and its centroid is $(0,0,0)$ and its third vertex $\mathrm{C}(4,7, \mathrm{c})$.

Using the formula, we get

$=\left(\frac{-2+4+a}{3}, \frac{1+7+b}{3}, \frac{3-5+c}{3}\right)$

$=\left(\frac{2+a}{3}, \frac{8+b}{3}, \frac{-2+c}{3}\right)$

Equating it with the coordinates of centroid, we get

$0=\frac{2+a}{3}$

$a=-2$

$0=\frac{8+b}{3}$

$b=-8$

and,

$\frac{-2+c}{3}=0$

c = 2

therefore, $a=-2, b=-8, c=2$.