Question

Find the equation of a line which passes through point $(1,2,3)$ and parallel to the line
$\frac{-x-2}{1}=\frac{y+3}{7}=\frac{2z-6}{3}$

Answer 1

Consider the problem ,

Let the required line is parallel to

$\begin{array}{c}\frac{-x-2}{1}=\frac{y+3}{7}=\frac{2z-6}{3}\\ \\ \frac{x+2}{-1}=\frac{y+3}{7}=\frac{z-\frac{6}{2}}{\frac{3}{2}}\end{array}$

Hence, its direction ratios are $-1,7,\frac{3}{2}$

So, direction ratios are same if they are parallel,

And passing through the point $(1,2,3)$

Therefore,

Equation is

$\frac{x-1}{-1}=\frac{y-2}{7}=\frac{z-3}{\frac{3}{2}}$.