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:\(\mathbf{r}_1 = \begin{pmatrix} 1+\lambda \\ 1 \\ 2+3\lambda \end{pmatrix}\), \(\mathbf{r}_2 = \begin{pmatrix} -2+\mu \\ 3-\mu \\ -1 \end{pmatrix}\), and \(\mathbf{r}_3 = \begin{pmatrix} 2-t \\ -1+2t \\ -1+3t \end{pmatrix}\)

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 \(A = \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}\)

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 \(B = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\)

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 \(C = \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}\)

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: \(A = \boxed{\sqrt{2}}\)