Given lines,
$$
y-3 k x+4=0
$$
$$
\text { and }(2 k-1) x-(8 k-1) y-6=0
$$
Slope of Eq. (i),
$$
\begin{array}{ll}
\text { Slope of Eq. (i), } & m_1=\frac{-(-3 k)}{1}=3 k \\
\text { and slope of Eq. (ii) } & m_2=\frac{-(2 k-1)}{-(8 k-1)}=\frac{2 k-1}{8 k-1}
\end{array}
$$
Since, lines are perpendicular.
$$
\begin{array}{rlrl}
& \therefore & m_1 m_2 & =-1 \\
\Rightarrow & & 3 k \times \frac{2 k-1}{8 k-1} & =-1 \\
\Rightarrow & 3 k \times(2 k-1) & =-(8 k-1) \\
\Rightarrow & 6 k^2-3 k & =-8 k+1 \\
\Rightarrow & & 6 k^2-3 k+8 k-1 & =0 \\
\Rightarrow & 6 k^2+5 k-1 & =0 \\
\Rightarrow & & 6 k^2+6 k-k-1 & =0 \\
\Rightarrow & & 6 k(k+1)-1(k+1) & =0 \\
\Rightarrow & & (6 k-1)(k+1) & =0
\end{array}
$$

$$
\therefore \quad k=\frac{1}{6},-1
$$