The equation of a line passing through the point $(2,4,6)$ and parallel to the line $3 x+4=4 y-1=1-4 z$ is
The equation of a line passing through the point $(2,4,6)$ and parallel to the line
$3 x+4=4 y-1=1-4 z$ is
$\frac{x-2}{4}=\frac{y-4}{3}=\frac{z-6}{3}$
$\frac{x-2}{4}=\frac{y-4}{3}=\frac{z-6}{-3}$
$\frac{x-2}{-4}=\frac{y-4}{3}=\frac{z-6}{-3}$
$\frac{x-2}{-4}=\frac{y-4}{-3}=\frac{z-6}{-3}$
Solution
Given equation of line is
$\frac{3 x+4}{x+\frac{4}{3}}=\frac{y-\frac{1}{4}}{\left(\frac{1}{3}\right)}=\frac{z-\frac{1}{4}}{\left(\frac{1}{4}\right)}=\frac{3\left(x+\frac{4}{3}\right)}{\left(-\frac{1}{4}\right)}=\frac{4\left(y-\frac{1}{4}\right)}{1}=\frac{-4\left(z-\frac{1}{4}\right)}{1}$
$\begin{aligned}
& \frac{x+\frac{4}{3}}{\left(\frac{1}{3}\right)}=\frac{y-\frac{1}{4}}{\left(\frac{1}{4}\right)}=\frac{z-\frac{1}{4}}{\left(-\frac{1}{4}\right)} \\
\therefore & \frac{1}{3}, \frac{1}{4},-\frac{1}{4} \text { are d.r. of a line i.e. } 4,3,-3
\end{aligned}$
Hence eq. of required line is
$\frac{x-2}{4}=\frac{y-4}{3}=\frac{z-6}{-3}$