Question: Consider the two lines $\ell\_1:\mathbf{r}=\begin{pmatrix}1\\1\\0\end{pmatrix}+s\begin{pmatrix}-1\\2\\-2\end{pmatrix}$ and $\ell\_2:\mathbf{r}=\begin{pmatrix}-2\\3\\-2\end{pmatrix}+t\begin{pmatrix}0\\-1\\2\end{pmatrix}$. I need to determine the equation of the line $\ell\_3$ whcich intersects both $\ell\_1$ and $\ell\_2$ at right angles.

I know that the direction of $\ell\_3$ will be given by $\begin{pmatrix}2\\2\\1\end{pmatrix}$, by the cross product of the two direction vectors of the first two lines. I am not sure how to proceed from here.

Answer: 
If
\begin{align}
\mathbf{r}\_1&=\begin{pmatrix}1\\1\\0\end{pmatrix}+t\_1\begin{pmatrix}-1\\2\\-2\end{pmatrix}\ \text{and}\\
\mathbf{r}\_2&=\begin{pmatrix}-2\\3\\-2\end{pmatrix}+t\_2\begin{pmatrix}0\\-1\\2\end{pmatrix}
\end{align}
are the points of intersection of $\ \ell\_1\ $ and $\ \ell\_2\ $ with $\ \ell\_3\ $, respectively, then $\ \mathbf{r}\_2-\mathbf{r}\_1\ $ must be in the same direction as $\ \ell\_3\ $. That is
\begin{align}
\mathbf{r}\_2-\mathbf{r}\_1&=\begin{pmatrix}-3\\2\\-2\end{pmatrix}-t\_1\begin{pmatrix}-1\\2\\-2\end{pmatrix}+t\_2\begin{pmatrix}0\\-1\\2\end{pmatrix}\\
&=t\_3\begin{pmatrix}2\\2\\1\end{pmatrix}\ .
\end{align}
for some $\ t\_3\ $. Can you see how to determine the values of $\ t\_1,t\_2\ $and $\ t\_3\ $?
Take a vector $\mathbf u$ from a point on $\ell\_1$ to a pont on $\ell\_2.$ Find its entries by subtraction. $$\mathbf u=[s-3,-t-2s+2,2t+2s-2]$$. A vector along $\ell\_1$ is $\mathbf v=[-1,2,-2]$. A vector along $\ell\_2$ is $\mathbf w=[0,-1,2]$. Then the two equations $\mathbf v \bullet \mathbf u=0$ and $\mathbf w \bullet \mathbf u=0$ give two equations for $s$ and $t$. Solve these eqations. Substitute the value of $s$ into the parametric equation of $\ell\_1.$ Substitute the value of $t$ into the parametric equation of $\ell\_2.$ You have found the points at which your desired line intersects the given lines.