Question

Find the equation of the line passing through the point $(-1,-1,2)$ and parallel to the line $2x-2=3y+1=6z-2$.

Answer 1

Let the given line

$2x-2=3y+1=6z-2$

$\begin{array}{c}2\left(x-1\right)=3\left(y+\frac{1}{3}\right)=6\left(z-\frac{1}{3}\right)\\ \\ \frac{x-1}{\left(\frac{1}{2}\right)}=\frac{y+\frac{1}{3}}{\left(\frac{1}{3}\right)}=\frac{z-\frac{1}{3}}{\left(\frac{1}{6}\right)}\end{array}$

Since, the required line is parallel to the given line

So, direction ratios proportional to

$\frac{1}{2},\frac{1}{3},\frac{1}{6}$

Therefore, required equation of line

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

$2x+2=3y+3=6z-12$

Which sis the required equation of line.