Given, $f(x)=a x^2+b x+c$ and GCD of $a, b, c$ is
1 .
Also given, $\frac{-7+\sqrt{1 \mathrm{l}} i}{6}$ is a root of $f(x)=0$, then $\frac{-7-\sqrt{11} i}{6}$ must be another root of $f(x)=0$.
So, sum of roots $=-\frac{b}{a}$
$$
\Rightarrow \quad \frac{-7+\sqrt{11} i-7-\sqrt{11} i}{6}=-\frac{b}{a}
$$


$\begin{aligned} & \text { and product of roots }=\frac{c}{a} \\ & \Rightarrow \quad\left(\frac{-7+\sqrt{11} i}{6}\right)\left(\frac{-7-\sqrt{11} i}{6}\right)=\frac{c}{a} \\ & \Rightarrow \quad \frac{49+11}{6}=\frac{c}{a}\end{aligned}$

From Eqs. (i) and (ii),
$$
\begin{aligned}
& a=3, b=7 \text { and } c=5 \text { and } \operatorname{GCD}(a, b, c)=1 \\
& \therefore f(x)=3 x^2+7 x+5
\end{aligned}
$$
$$
\begin{aligned}
& \text { Now, } f\left(\frac{x}{k}\right)-L=(x+4)(3 x-5) \\
& \Rightarrow 3\left(\frac{x}{k}\right)^2+7\left(\frac{x}{k}\right)+5-L=3 x^2+7 x-20 \\
& \Rightarrow 3 x^2+7 k x+5 k^2-L k^2=3 k^2 x^2+7 k^2 x-20 k^2
\end{aligned}
$$
On comparing the coefficients, we get

From Eqs. (iii) and (iv),
$k=1$ and $L=25$