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

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
\(\frac{x-2}{4}=\frac{y-4}{3}=\frac{z-6}{-3}\)