6 Show that the lines with equations $r=\left(\begin{array}{l}1\\ 1\\ 2\end{array}\right)+\lambda \left(\begin{array}{l}1\\ 0\\ 3\end{array}\right),\mathbf{r}=\left(\begin{array}{c}-2\\ 3\\ -1\end{array}\right)+\mu \left(\begin{array}{c}1\\ -1\\ 0\end{array}\right)$ and $\mathbf{r}=\left(\begin{array}{c}2\\ -1\\ -1\end{array}\right)+t\left(\begin{array}{c}-1\\ 2\\ 3\end{array}\right)$ form a triangle, and find its area.

Step 1 ：First, we write the parametric equations of the lines as:${\mathbf{r}}_1=\left(\begin{array}{c}1+\lambda \\ 1\\ 2+3\lambda \end{array}\right)$, ${\mathbf{r}}_2=\left(\begin{array}{c}-2+\mu \\ 3-\mu \\ -1\end{array}\right)$, and ${\mathbf{r}}_3=\left(\begin{array}{c}2-t\\ -1+2t\\ -1+3t\end{array}\right)$

Step 2 ：Next, we find the intersection points of the lines by setting the corresponding components equal. For ${\mathbf{r}}_1$ and ${\mathbf{r}}_2$, we have $1+\lambda =-2+\mu$ and $2+3\lambda =-1$. Solving for $\lambda$ and $\mu$, we get $\lambda =-1$ and $\mu =0$. Thus, the intersection point is $A=\left(\begin{array}{c}0\\ 1\\ -1\end{array}\right)$

Step 3 ：Similarly, for ${\mathbf{r}}_2$ and ${\mathbf{r}}_3$, we have $-2+\mu =2-t$ and $3-\mu =-1+2t$. Solving for $t$ and $\mu$, we get $t=1$ and $\mu =2$. Thus, the intersection point is $B=\left(\begin{array}{c}0\\ 1\\ 1\end{array}\right)$

Step 4 ：Finally, for ${\mathbf{r}}_1$ and ${\mathbf{r}}_3$, we have $1+\lambda =2-t$ and $2+3\lambda =-1+3t$. Solving for $t$ and $\lambda$, we get $t=2$ and $\lambda =0$. Thus, the intersection point is $C=\left(\begin{array}{c}1\\ 1\\ 2\end{array}\right)$

Step 5 ：Now that we have the vertices of the triangle, we can find the side lengths. We have $AB=\sqrt{(0-0{)}^2+(1-1{)}^2+(-1-1{)}^2}=\sqrt{4}=2$, $BC=\sqrt{(1-0{)}^2+(1-1{)}^2+(2-1{)}^2}=\sqrt{2}$, and $CA=\sqrt{(1-0{)}^2+(1-1{)}^2+(2+1{)}^2}=\sqrt{10}$

Step 6 ：Using Heron's formula, we first find the semi-perimeter: $s=\frac{2+\sqrt{2}+\sqrt{10}}{2}$

Step 7 ：Then, we find the area of the triangle: $A=\sqrt{s(s-2)(s-\sqrt{2})(s-\sqrt{10})}$

Step 8 ：Calculating the area, we get $A=\sqrt{\frac{1}{2}(2+\sqrt{2}+\sqrt{10})(\sqrt{2}+\sqrt{10})(\sqrt{10}-2)(\sqrt{2}-2)}$

Step 9 ：Finally, we simplify the expression to get the area of the triangle: $A=\boxed{{\displaystyle \sqrt{2}}}$