**Question:**

Find the Cartesian equation of the line which passes through the point

$(-2,4,-5)$ and parallel to the line given by $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$

**Solution:**

It is given that the line passes through the point $(-2,4,-5)$ and is parallel to $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$

The direction ratios of the line, $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$, are 3,5, and 6 .

The required line is parallel to $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$

Therefore, its direction ratios are $3 k, 5 k$, and $6 k$, where $k \neq 0$

It is known that the equation of the line through the point $\left(x_{1}, y_{1}, z_{1}\right)$ and with direction ratios, $a, b, c$, is given by $\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}$

Therefore the equation of the required line is

$\frac{x+2}{3 k}=\frac{y-4}{5 k}=\frac{z+5}{6 k}$

$\Rightarrow \frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}=k$