Step 1: Problem Identification: The objective is to compute the shortest distance between two lines presented in Cartesian form. The initial step involves categorizing the lines as parallel, intersecting, or skew.Step 2: Theoretical Foundation: A line in Cartesian form is represented as $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$, which corresponds to the vector form $\vec{r} = (x_1\hat{i} + y_1\hat{j} + z_1\hat{k}) + \lambda(a\hat{i} + b\hat{j} + c\hat{k})$. The direction ratios for the first line are $(a_1, b_1, c_1)$ and for the second line are $(a_2, b_2, c_2)$. Lines are parallel if their direction ratios are proportional, i.e., $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$. For parallel lines, the shortest distance is the perpendicular distance from a point on one line to the other. If lines are not parallel, they are either intersecting or skew.Step 3: Calculation and Verification:Consider line L1: $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$, with direction ratios $\vec{d_1} = (2, 3, 4)$.Consider line L2: $\frac{x-2}{4} = \frac{y-4}{6} = \frac{z-5}{8}$, with direction ratios $\vec{d_2} = (4, 6, 8)$.Parallelism check: $\frac{4}{2} = 2$, $\frac{6}{3} = 2$, $\frac{8}{4} = 2$. Since the ratios are equal, $\vec{d_2} = 2\vec{d_1}$, confirming the lines are parallel.To distinguish between coincident and distinct parallel lines, test a point from L1 on L2. A point on L1 is $P_1 = (1, 2, 3)$. Substituting into L2's equation:\[ \frac{1-2}{4} = \frac{-1}{4} \]\[ \frac{2-4}{6} = \frac{-2}{6} = \frac{-1}{3} \]\[ \frac{3-5}{8} = \frac{-2}{8} = \frac{-1}{4} \]As $\frac{-1}{4} eq \frac{-1}{3}$, $P_1$ is not on L2. The lines are distinct and parallel.The shortest distance $D$ between two parallel lines $\vec{r} = \vec{a_1} + \lambda\vec{d}$ and $\vec{r} = \vec{a_2} + \mu\vec{d}$ is given by $D = \frac{|(\vec{a_2} - \vec{a_1}) \times \vec{d}|}{|\vec{d}|}$.From the line equations:$\vec{a_1} = \hat{i} + 2\hat{j} + 3\hat{k}$$\vec{a_2} = 2\hat{i} + 4\hat{j} + 5\hat{k}$$\vec{d} = 2\hat{i} + 3\hat{j} + 4\hat{k}$$\vec{a_2} - \vec{a_1} = (2-1)\hat{i} + (4-2)\hat{j} + (5-3)\hat{k} = \hat{i} + 2\hat{j} + 2\hat{k}$Cross product: \[ (\vec{a_2} - \vec{a_1}) \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 2 & 3 & 4 \end{vmatrix} = \hat{i}(8-6) - \hat{j}(4-4) + \hat{k}(3-4) = 2\hat{i} - \hat{k} \]Magnitude of cross product: $|2\hat{i} - \hat{k}| = \sqrt{2^2 + (-1)^2} = \sqrt{5}$.Magnitude of direction vector: $|\vec{d}| = \sqrt{2^2 + 3^2 + 4^2} = \sqrt{4+9+16} = \sqrt{29}$.Distance $D = \frac{\sqrt{5}}{\sqrt{29}} = \sqrt{\frac{5}{29}}$.