Answer
$$\operatorname{proj}_{{v}} {u}
=\frac{6}{29}(3,2,4).$$
Work Step by Step
Let $u=(0,-1,2)$, $v=(3,2,4)$, $\langle{u}, {v}\rangle=u\cdot v$. Then, we have
$$\langle{u}, {v}\rangle=6, \quad \langle{v}, {v}\rangle=29.$$
Now, the orthogonal projection of $u$ onto $v$ is given by
$$\operatorname{proj}_{{v}} {u}
=\frac{\langle{u}, {v}\rangle}{\langle{v}, {v}\rangle} {v}=\frac{6}{29}(3,2,4).$$