Two vertices of a triangle ABC are A

Solution:

Since the centroid of a triangle

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

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

Using the formula, we get

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

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

Equating it with the coordinates of centroid, we get

$1=\frac{5+a}{3}$

$a=-2$

$0=\frac{-5+b}{3}$

$b=-5$

and,

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

c = 7

therefore, the point is $(-2,-5,7)$