Equation of a line orthogonal to two other lines 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.
 A: 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\ $?
A: 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.
A: 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.
