Find the direction cosines of the vector joining the points

The given points are $A(1,2,-3)$ and $B(-1,-2,1)$.

$\therefore \overrightarrow{\mathrm{AB}}=(-1-1) \hat{i}+(-2-2) \hat{j}+\{1-(-3)\} \hat{k}$

$\Rightarrow \overrightarrow{\mathrm{AB}}=-2 \hat{i}-4 \hat{j}+4 \hat{k}$

$\therefore|\overrightarrow{\mathrm{AB}}|=\sqrt{(-2)^{2}+(-4)^{2}+4^{2}}=\sqrt{4+16+16}=\sqrt{36}=6$

Hence, the direction cosines of $\overrightarrow{\mathrm{AB}}$ are $\left(-\frac{2}{6},-\frac{4}{6}, \frac{4}{6}\right)=\left(-\frac{1}{3},-\frac{2}{3}, \frac{2}{3}\right)$.