Lines in 3D

Try to complete the following questions for Monday’s lesson.

1. Find the distance of the point  $A(-1,1,2)$ to the line $$\displaystyle{\vec{r}=\begin{bmatrix}1\\-2\\3\end{bmatrix}+\lambda \begin{bmatrix}-2\\1\\1\end{bmatrix}}$$
2. The lines below intersect at a point $A$. Use an algebraic method to find the coordinates of  $A$, then verify your answer using GeoGebra.
   $$L_1: \vec{r}=\begin{bmatrix}-1\\-1\\4\end{bmatrix}+\lambda \begin{bmatrix}-2\\1\\1\end{bmatrix}$$ $$L_2: \vec{r}=\begin{bmatrix}-2\\0\\7\end{bmatrix}+\lambda \begin{bmatrix}-6\\4\\8\end{bmatrix}$$
3. In 2D, any two non-parallel lines will have a point of intersection. In 3D, two lines can be non-parallel and have no point of intersection; such lines are called skew lines.
   Show that $L_1$ from question 2 and $L_3$ below are skew lines.
   $$L_3: \vec{r}=\begin{bmatrix}-2\\5\\12\end{bmatrix}+\lambda \begin{bmatrix}-6\\4\\8\end{bmatrix}$$