Chapter 13 - Vector Geometry - Chapter Review Exercises - Page 701: 13

$$ u=\frac{1}{6}\langle 2,-11,7 \rangle.$$

Since $ w=\langle 2,-2,1 \rangle $ and $ v=\langle4,5,-4 \rangle $, we have \begin{align*}v+5u =3w-u&\Longrightarrow 6u=3w-v \\ &\Longrightarrow u=\frac{1}{6}(3w-v)\\ &\Longrightarrow u=\frac{1}{6}(3\langle 2,-2,1 \rangle-\langle 4,5,-4 \rangle)\\ &\Longrightarrow u=\frac{1}{6}\langle 2,-11,7 \rangle. \end{align*}