Answer
See the details below.
Work Step by Step
Since $r(t)=\lt t,1,1 \gt$, then $r'(t)=\lt 1,0,0 \gt$.
$$\|r(t)\|=\sqrt{2+t^2}, \quad \|r'(t)\|=1$$
and moreover $$\|r(t)\|'=\frac{d}{dt} \|r(t)\|=\frac{t}{\sqrt{2+t^2}}.$$
That is $$\|r(t)\|'\neq \|r'(t)\|.$$
