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: 
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.
Hint:

Take an arbitrary $u$-point on the first line $L\_1$:
$$\frac{x-1}{-1}=\frac{y-1}{2}=\frac{z-0}{-2}=u \implies x=-u+1,y=2u+1,z=-2u$$
Take a general $v$-point on the second line L\_2:
$$\frac{x+2}{0}=\frac{y-3}{-1}=\frac{z+2}{3}=v \implies x=-2,y=3-v,z=3v-2$$
Take $\vec{UV}$, demand $\vec{UV}.\vec{L\_1}=0=\vec{UV}.\vec{L\_2}$, determine $u,v$. Find position vectors $\vec U$, $\vec{V}$. Find equation of the line UV which is nothing but $L\_3$.

**EDIT**
Since you know $\vec{L\_3}=\vec{L\_1}\times \vec{L\_2}$, you may also set $\vec{UV}=p\vec{L\_3}$ to proceed.