Let $L_1, L_2$ be the lines represented by the equation $4x^2-5xy+3y^2=0$. Let $L_3, L_4$ be two lines passing through the point $\left(4,3\right)$ such that $L_3$ and $L_4$ are perpendicular to $L_1$ and $L_2$ respectively. If the combined equation of $L_3$ and $L_4$ is $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$, then $af+bg+ch=$

Let $L_1, L_2$ be the lines represented by the equation $4x^2-5xy+3y^2=0$,

Now comparing with $a{x}^2+2hxy+b{y}^2=0$ we get,

$m_1+m_2=\frac{-2h}{b}=\frac{5}{3}$ and $m_1{m}_2=\frac{a}{b}=\frac{4}{3}$

Now let slope of perpendicular line $L_3, L_4$ be  $\frac{-1}{m_1} \& \frac{-1}{m_2}$ which is passing through the point $\left(4,3\right)$ 

So, equation of pair will be ${\left(y-3\right)}^2-\left(\frac{-1}{m_1}+\frac{-1}{m_2}\right)\left(x-4\right)\left(y-3\right)+\frac{1}{m_1{m}_2}{\left(x-4\right)}^2=0$

$\Rightarrow {\left(y-3\right)}^2+\left(\frac{1}{m_1}+\frac{1}{m_2}\right)\left(x-4\right)\left(y-3\right)+\frac{1}{m_1{m}_2}{\left(x-4\right)}^2=0$

$\Rightarrow {\left(y-3\right)}^2+\left(\frac{5}{4}\right)\left(x-4\right)\left(y-3\right)+\frac{3}{4}{\left(x-4\right)}^2=0$

$\Rightarrow 3x^2+4y^2+5xy-39x-44y+144=0$

Now on comparing with $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$,

We get the value of $af+bg+ch=216$