Step 1

First of all, the derivative of x is

\(\displaystyle{x}'{\left({t}\right)}={\left(-{3},-{1},-{1}\right)}\)

Its norm is

\(\mid x'\left(t\right)\mid =\sqrt{\left(-3\right)^{2} +\left(-1\right)^{2} +\left(-1\right)^{2} }= \sqrt{9+1+1}= \sqrt{11}\)

Now we find

\(s\left(t\right)=\int_{t_{0}}^{t} \mid x'\left(u\right) \mid du\),

where \(\displaystyle{x}{\left({t}_{{{0}}}\right)}\) is the starting point So, we must have that

\(\displaystyle{x}{\left({t}_{{{0}}}\right)}={\left({3},−{2},−{1}\right)}\)

which yields

\(\displaystyle{t}_{{{0}}}={0}\)

Therefore,

\(s\left(t\right)=\int_{0}^{t} \mid x'\left(u\right) \mid du=\int_{0}^{t}\sqrt{11}du=\sqrt{11}t\)

Therelore, we set

\(\displaystyle{t}={\frac{{{s}}}{{\sqrt{{{11}}}}}}\)

This means that the reparametrization of x in terms of arclength is

\(\widetilde{x}\left(s\right)=x\left(\frac{s}{\sqrt{11}}\right)=\left(3-\frac{3s}{\sqrt{11}}, -2-\frac{s}{\sqrt{11}}, -1-\frac{s}{\sqrt{11}}\right)\)

To write it clearly,

\(\widetilde{x}\left(s\right)=\left(3-\frac{3s}{\sqrt{11}}, -2-\frac{s}{\sqrt{11}}, -1-\frac{s}{\sqrt{11}}\right)\)