Given Equation 9x²+12xy+ky²=0 –(1) and 2x+3y – 7 = 0 –(2)

m=slope= -2/3

If (2) is makes equal angles with (1) then (2) is parallel to Angle bisector of (1)

Angle bisector of (1)

$\begin{array}{l}h\left(x^2-y^2\right)=\left(a-b\right)xy\\ \Rightarrow 6\left(1-{\left(\frac{y}{x}\right)}^2\right)=\left(9-k\right)\left(\frac{y}{x}\right)\\ \Rightarrow 6\left(1-m^2\right)=\left(9-k\right)m\\ \Rightarrow 6\left(1-\frac{4}{9}\right)=\left(9-k\right)\left(\frac{-2}{3}\right)\\ \Rightarrow 6\left(\frac{5}{9}\right)\times \left(\frac{-3}{2}\right)=9-k\\ \Rightarrow -5=9-k\\ \Rightarrow k=9+5=14\end{array}$