We know that,
If two lines \(l_1 x+m_1 y+n_1=0\) and
\(l_2 x+m_2 y+n_2=0\) are conjugate lines with respect to the hyperbola
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \text {, then } a^2 l_1 l_2-b^2 m_1 m_2=n_1 n_2\)
For hyperbola,
\(5 x^2-6 y^2=15 \Rightarrow \frac{x^2}{3}-\frac{y^2}{5 / 2}=1\)
Here, \(a^2=3, b^2=\frac{5}{2}, l_1=2, l_2=3\),
and
\(m_1=-k, m_2=-1\)
\(\begin{array}{lc}
\text {and } & n_1=3, n_2=1 \\
\therefore & 3 \times 2 \times 3-\frac{5}{2}(-k)(-1)=3 \times 1 \\
\Rightarrow & 18-\frac{5}{2} k=3 \Rightarrow 15=\frac{5}{2} k \Rightarrow k=6
\end{array}\)