1. **Problem Statement:** Determine if the two metro tracks (Line A and Line B) are parallel. Find the equation of a line parallel to Line A's track $l_1$ passing through point $(1, -2, -3)$. Find the equation of a line passing through point $(3, 2, 1)$ for power supply setup. 2. **Given Lines:** Line $l_1$: $\frac{x-2}{3} = \frac{y+1}{2} = \frac{z-3}{-1}$ Line $l_2$: $\frac{x-1}{2} = \frac{y-3}{1} = \frac{z+2}{-3}$ Line $l_3$: Given as $1 - \frac{1}{2} = 3$ which is incomplete or unclear, so we consider only $l_1$ and $l_2$ for parallelism. 3. **Step 1: Check if $l_1$ and $l_2$ are parallel.** - Direction vector of $l_1$ is $\vec{d_1} = (3, 2, -1)$. - Direction vector of $l_2$ is $\vec{d_2} = (2, 1, -3)$. Two lines are parallel if their direction vectors are scalar multiples: Check if $\vec{d_1} = k \vec{d_2}$ for some scalar $k$. Compare components: $\frac{3}{2} = 1.5$, $\frac{2}{1} = 2$, $\frac{-1}{-3} = \frac{1}{3} \approx 0.333$. Since these ratios are not equal, $l_1$ and $l_2$ are **not parallel**. 4. **Step 2: Equation of line parallel to $l_1$ passing through $(1, -2, -3)$.** - Use point-direction form: $$\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c}$$ where $(x_0, y_0, z_0) = (1, -2, -3)$ and direction vector $\vec{d_1} = (3, 2, -1)$. So, $$\frac{x - 1}{3} = \frac{y + 2}{2} = \frac{z + 3}{-1}$$ 5. **Step 3: Equation of line passing through $(3, 2, 1)$ for power supply setup.** - The problem does not specify direction, so we assume it is parallel to $l_1$ as well (common in such contexts). Using the same direction vector $\vec{d_1} = (3, 2, -1)$ and point $(3, 2, 1)$: $$\frac{x - 3}{3} = \frac{y - 2}{2} = \frac{z - 1}{-1}$$ **Final answers:** (i) Lines $l_1$ and $l_2$ are **not parallel**. (ii) Equation of line parallel to $l_1$ through $(1, -2, -3)$: $$\frac{x - 1}{3} = \frac{y + 2}{2} = \frac{z + 3}{-1}$$ (iii) Equation of line through $(3, 2, 1)$ for power supply: $$\frac{x - 3}{3} = \frac{y - 2}{2} = \frac{z - 1}{-1}$$