We have,
$4 x^2+6 x y+k y^2=0$
and $\quad 3 x^2-5 x y+2 y^2=0$
Let $y=m x$ be one of the lines represented by Eq. (i). Then, $y=m x$ satisfies Eq. (i)
$\begin{aligned} & \therefore \quad 4 x^2+6 x \cdot m x+k m^2 n^2=0 \\ & \Rightarrow \quad k m^2+6 m+4=0 \\ & \end{aligned}$
The equation of a line passing through the origin and perpendicular to $y=m x$, is
$y-0=-\frac{1}{m}(x-0) \Rightarrow m y+x=0$
Since, one of the lines given by Eq. (i) is perpendicular to one of the lines given by $3 x^2-5 x y+2 y^2=0$. Therefore one of the lines given by (ii) is $m y+x=0$
Hence, $m y+x=0$ i.e., $x=-m y$ satisfies Eq. (ii).
$\begin{array}{rrrl}\therefore & 3(-m y)^2-5 y(-m y)+2 y^2=0 \\ \Rightarrow & 3 m^2 y^2+5 m y^2+2 y^2=0 \\ \Rightarrow & 3 m^2+5 m+2=0\end{array}$
using Eqs. (iii) and (iv), we get
$\begin{aligned} & \Rightarrow(4 \times 3-k \times 2)^2+4\left(3 \times 3+\left(\frac{-5}{2}\right) \times k\right) \\ & \quad\left(3 \times 2+\left(\frac{-5}{2}\right) \times 4\right)=0 \\ & \Rightarrow(12-2 k)^2+4\left(9-\frac{5 k}{2}\right)(6-10)=0 \\ & \Rightarrow \quad(12-2 k)^2-16\left(9-\frac{5 k}{2}\right)=0 \\ & \Rightarrow \quad 144+4 k^2-48 k-144+40 k=0 \\ & \Rightarrow \quad 4 k^2-8 k=0 \Rightarrow 4 k^2=8 k \\ & \Rightarrow \quad k=2\end{aligned}$