The Cartesian equation of a line $AB$ is: $\frac{2x - 1}{12} = \frac{y + 2}{2} = \frac{z - 3}{3}$. Find the direction cosines of a line parallel to line $AB$.

$(\frac{6}{7}, \frac{2}{7}, \frac{3}{7})$

The correct answer is Option (2) → $(\frac{6}{7}, \frac{2}{7}, \frac{3}{7})$ ##

We have,

$\frac{2x - 1}{12} = \frac{y + 2}{2} = \frac{z - 3}{3}$

The equation of line $AB$ can be rewritten as:

$\frac{x - \frac{1}{2}}{6} = \frac{y - (-2)}{2} = \frac{z - 3}{3}$

Thus, direction ratios of the line parallel to $AB$ are proportional to $6, 2, 3$.

Hence, the direction cosines of the line parallel to AB are proportional to

$\left( \frac{6}{\sqrt{6^2 + 2^2 + 3^2}}, \frac{2}{\sqrt{6^2 + 2^2 + 3^2}}, \frac{3}{\sqrt{6^2 + 2^2 + 3^2}} \right)$

$= \left( \frac{6}{\sqrt{49}}, \frac{2}{\sqrt{49}}, \frac{3}{\sqrt{49}} \right)$

$\text{ or } \left( \frac{6}{7}, \frac{2}{7}, \frac{3}{7} \right)$