$$
\begin{aligned}
& \text { } C_1: x^2+y^2+x+y-4=0 \\
& \quad C_2: 3 x^2+3 y^2-x-y-\lambda=0 \\
& \Rightarrow \quad x^2+y^2-\frac{x}{3}-\frac{y}{3}-\frac{\lambda}{3}=0
\end{aligned}
$$
$\therefore$ Length of tangent drawn by an external point is $\sqrt{S_1}$.
$\therefore$ Length of tangent drawn from $(1,2)$ to $C_1$ and $C_2$ are $L_1$ and $L_2$ respectively.
$$
\begin{aligned}
L_1 & =\sqrt{S_1}=\sqrt{1^2+2^2+1+2-4} \\
& =\sqrt{1+4+1+2-4}=\sqrt{4} \\
L_1 & =2
\end{aligned}
$$
$$
\therefore L_2=\sqrt{S_1^{\prime}}=\sqrt{1^2+2^2-\frac{1}{3}-\frac{2}{3}-\frac{\lambda}{3}}=\sqrt{4-\frac{\lambda}{3}}
$$
Given that,
$$
\frac{L_1}{L_2}=\frac{3}{4}
$$
$\begin{aligned} &=\frac{2}{\sqrt{4-\frac{\lambda}{3}}}=\frac{3}{4} \Rightarrow \sqrt[3]{4-\frac{\lambda}{3}}=8 \\ & \Rightarrow \quad 9\left(4-\frac{\lambda}{3}\right)=64 \Rightarrow 4-\frac{\lambda}{3}=\frac{64}{9} \\ & \Rightarrow \quad 4-\frac{64}{9}=\frac{\lambda}{3} \Rightarrow \frac{36-64}{9}=\frac{\lambda}{3} \\ & \lambda=-\frac{28}{3}\end{aligned}$