Two vertices of $\triangle \mathrm{ABC}$ are $\mathrm{B}(-3,1)$ and $\mathrm{C}(0,-2)$. Let the third vertex be $\mathrm{A}(\mathrm{a}, \mathrm{b})$.

=>  the coordinates of its centroid are

$\left(\frac{-3+0+a}{3}, \frac{1-2+b}{3}\right)$

i.e., $\left(\frac{-3+a}{3}, \frac{-1+b}{3}\right)$

Since,  it is given that the centroid is at the origin, that is $\mathrm{G}(0,0)$. Therefore

$0=\frac{-3+a}{3}, 0=\frac{-1+b}{3}$

$\Rightarrow 0=-3+a, 0=-1+b$

$\Rightarrow 3=a, 1=b$

$\Rightarrow a=3, b=1$

Therefore, the third vertex of $\triangle \mathrm{ABC}$ is $\mathrm{A}(3,1)$.