1. **State the problem:** We are given two lines: $$\mathbf{r}_1 = (1, -1, 1) + t(3, 2, 1), t \in \mathbb{R}$$ $$\mathbf{r}_2 = (-2, -3, 0) + s(1, 2, 3), s \in \mathbb{R}$$ We need to: a. Find the coordinates of their point of intersection. b. Find a vector equation for the line perpendicular to both given lines and passing through their point of intersection. --- 2. **Find the point of intersection (part a):** The lines intersect if there exist parameters $t$ and $s$ such that: $$ (1 + 3t, -1 + 2t, 1 + t) = (-2 + s, -3 + 2s, 0 + 3s) $$ This gives the system: $$\begin{cases} 1 + 3t = -2 + s \\ -1 + 2t = -3 + 2s \\ 1 + t = 3s \end{cases}$$ 3. **Solve the system:** From the first equation: $$ s = 1 + 3t + 2 = 3 + 3t $$ From the third equation: $$ 1 + t = 3s \implies 1 + t = 3(3 + 3t) = 9 + 9t $$ Rearranged: $$ 1 + t = 9 + 9t \implies 1 - 9 = 9t - t \implies -8 = 8t \implies t = -1 $$ 4. **Find $s$ using $t = -1$: ** $$ s = 3 + 3(-1) = 3 - 3 = 0 $$ 5. **Check the second equation:** $$ -1 + 2(-1) = -3 + 2(0) \implies -1 - 2 = -3 + 0 \implies -3 = -3 $$ True, so the lines intersect at $t = -1$, $s = 0$. 6. **Find the intersection point coordinates:** Using $t = -1$ in $\mathbf{r}_1$: $$ (1 + 3(-1), -1 + 2(-1), 1 + (-1)) = (1 - 3, -1 - 2, 0) = (-2, -3, 0) $$ So the point of intersection is: $$ \boxed{(-2, -3, 0)} $$ --- 7. **Find the vector equation of the line perpendicular to both lines passing through the intersection (part b):** The direction vectors of the lines are: $$ \mathbf{v}_1 = (3, 2, 1), \quad \mathbf{v}_2 = (1, 2, 3) $$ The vector perpendicular to both is the cross product: $$ \mathbf{n} = \mathbf{v}_1 \times \mathbf{v}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 2 & 1 \\ 1 & 2 & 3 \end{vmatrix} $$ Calculate: $$ \mathbf{n} = (2 \cdot 3 - 1 \cdot 2, -(3 \cdot 3 - 1 \cdot 1), 3 \cdot 2 - 2 \cdot 1) = (6 - 2, -(9 - 1), 6 - 2) = (4, -8, 4) $$ Simplify by dividing by 4: $$ \mathbf{n} = (1, -2, 1) $$ 8. **Write the vector equation of the line:** Passing through $(-2, -3, 0)$ with direction $\mathbf{n}$: $$ \mathbf{r} = (-2, -3, 0) + u(1, -2, 1), u \in \mathbb{R} $$ --- **Final answers:** - Point of intersection: $$ \boxed{(-2, -3, 0)} $$ - Vector equation of the perpendicular line: $$ \boxed{\mathbf{r} = (-2, -3, 0) + u(1, -2, 1), u \in \mathbb{R}} $$