An alternative route is to first find the unit vector along $v$ and then take the scalar (signed) projection of $u$ along it.
$\|v\| = \sqrt{1^2+2^2+(-1)^2} = \sqrt{6}$, so the unit vector is $\hat v = \dfrac{1}{\sqrt6}(1,2,-1)$.
The scalar projection of $u$ along $\hat v$ is
\[u\cdot\hat v = \frac{1}{\sqrt6}(u\cdot v) = \frac{1}{\sqrt6}\big(2(1)+(-1)(2)+3(-1)\big) = \frac{-3}{\sqrt6} = -\frac{\sqrt6}{2}.\]
The vector projection is this scalar times the unit vector:
\[\text{proj}_v u = \left(-\frac{\sqrt6}{2}\right)\hat v = \left(-\frac{\sqrt6}{2}\right)\cdot\frac{1}{\sqrt6}(1,2,-1) = -\frac12(1,2,-1) = \left(-\frac12,\,-1,\,\frac12\right),\]
which agrees with the direct formula computation.
\[\boxed{\left(-\frac{1}{2},\,-1,\,\frac{1}{2}\right)}\]