Given that,
$$
\begin{aligned}
x & =\frac{1 \cdot 3}{3 \cdot 6}+\frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9}+\frac{1 \cdot 3 \cdot 5 \cdot 7}{3 \cdot 6 \cdot 9 \cdot 12}+\ldots \infty \text { terms } \\
\Rightarrow \quad x & =\frac{1 \cdot 3}{3^2(2 !)}+\frac{1 \cdot 3 \cdot 5}{3^3(3 !)}+\frac{1 \cdot 3 \cdot 5 \cdot 7}{3^4(4 !)}+\ldots \infty \text { terms }
\end{aligned}
$$
$$
\begin{aligned}
\Rightarrow \quad x= & \frac{\frac{1}{2}\left(\frac{1}{2}+1\right)}{2 !}\left(\frac{2}{3}\right)^2 \\
& +\frac{\left(\frac{1}{2}\right)\left(\frac{1}{2}+1\right)\left(\frac{1}{2}+2\right)}{3 !}\left(\frac{2}{3}\right)^3+\ldots \\
\Rightarrow \quad x= & {\left[1+\frac{1}{2}\left(\frac{2}{3}\right)+\frac{\frac{1}{2}\left(\frac{1}{2}+1\right)}{2 !}\left(\frac{2}{3}\right)^2\right.} \\
& \left.+\frac{\frac{1}{2}\left(\frac{1}{2}+1\right)\left(\frac{1}{2}+1\right)}{3 !}\left(\frac{2}{3}\right)^3 \ldots .\right]-\left(1+\frac{1}{3}\right)
\end{aligned}
$$
Now, by using binomial expansion for any index
$$
\begin{aligned}
& \Rightarrow x=\left(1-\frac{2}{3}\right)^{-1 / 2}-\frac{4}{3} \Rightarrow x=\left(\frac{1}{3}\right)^{-1 / 2}-\frac{4}{3} \\
& \Rightarrow \quad x=\sqrt{3}-\frac{4}{3} \Rightarrow 3 x+4=3 \sqrt{3}
\end{aligned}
$$
On squaring both the sides, we get
$$
\begin{aligned}
& \Rightarrow \quad(3 x+4)^2=3(\sqrt{3})^2 \Rightarrow 9 x^2+24 x+16=27 \\
& \Rightarrow \quad 9 x^2+24 x=11
\end{aligned}
$$