## Elementary Linear Algebra 7th Edition

$c=\pm \frac{1}{\sqrt{14}}$.
Since $\|c(1,2,3)\|=1$ and $c(1,2,3)=(c,2c,3c)$, we have $$\sqrt{c^2+4c^2+9c^2}=1 \Longrightarrow 14 c^2=1.$$ Hence, $c=\pm \frac{1}{\sqrt{14}}$.