3D Line Equations - Calculus 3

A line $L$ has the property that it passes through the point $(1, 2, 3)$ and is perpendicular to both the line $\frac{x-1}{2} = \frac{y}{1} = \frac{z+1}{-1}$ and the line $\vec{r} = \langle 0, 1, 2 \rangle + s\langle 1, -1, 1 \rangle$ . Which equation represents line $L$ ?
Answer: $\frac{x-1}{0} = \frac{y-2}{-3} = \frac{z-3}{-3}$ To find line $L$ , we need its direction vector, which must be perpendicular to both given lines. The first line has direction vector $\vec{v_1} = \langle 2, 1, -1 \rangle$ , and the second line has direction vector $\vec{v_2} = \langle 1, -1, 1 \rangle$ . The direction vector of $L$ is $\vec{v_1} \times \vec{v_2} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & 1 & -1 \\ 1 & -1 & 1 \end{vmatrix} = \vec{i}(1 \cdot 1 - (-1)(-1)) - \vec{j}(2 \cdot 1 - (-1)(1)) + \vec{k}(2(-1) - 1 \cdot 1) = \vec{i}(0) - \vec{j}(3) + \vec{k}(-3) = \langle 0, -3, -3 \rangle$ . Since line $L$ passes through $(1, 2, 3)$ with direction vector $\langle 0, -3, -3 \rangle$ , its symmetric equation is $\frac{x-1}{0} = \frac{y-2}{-3} = \frac{z-3}{-3}$ . Note that $\frac{x-1}{0}$ means the $x$ -coordinate is constant at $x = 1$ .