Q.Find the vector equation of the line whose Cartesian equation is $\dfrac{x-3}{2}=\dfrac{y+2}{-3}=\dfrac{z-5}{6}$.

Comparing $\dfrac{x-3}{2}=\dfrac{y-(-2)}{-3}=\dfrac{z-5}{6}$ with $\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}$: the base point is $(3,-2,5)$ and the direction ratios are $(2,-3,6)$. …

Match the given symmetric-form equation term-by-term against the general Cartesian form to read off the point and directi …

Reading the base point's $y$-coordinate as $+2$ instead of $-2$ (the sign flip inside $y-(-2)=y+2$ is easy to misread in reverse); keeping the direction ratios but forgetting to convert them to unit …