Answer
$ \lt \dfrac{-1}{3}, \dfrac{-1}{3},\dfrac{-1}{3}\gt$
Work Step by Step
The vector projections of $u$ onto $v$ is defined as: $proj_{v}u=(\dfrac{u \cdot v}{v \cdot v})v$
Thus, $proj_{v}u=[\dfrac{1(1)+(-2)(1)+(0)(-1)}{2(2)+1(1)+(1)(1)}]\lt 1,1,1 \gt=(\dfrac{-1}{3})\lt 1,1,1 \gt=\lt \dfrac{-1}{3}, \dfrac{-1}{3},\dfrac{-1}{3}\gt$
