12. Given that the center of ellipse $C$ is at the origin, the foci are on the $x$-axis, the eccentricity is $\frac{\sqrt{3}}{2}$, and the area of the triangle formed by any three vertices of ellipse $C$ is $\frac{1}{2}$.
(1) Find the equation of ellipse $C$;
(2) If a line $l$ passing through $P(\lambda, 0)$ intersects the ellipse at two distinct points $A, B$, and $\overrightarrow{A P}=2 \overrightarrow{P B}$, find the range of the real number $\lambda$.
【Analysis】: (1) Let the semi-major axis of the ellipse be $a$, and the semi-minor axis be $b$, then we have $\frac{\sqrt{a^{2}-b^{2}}}{a}=\frac{\sqrt{3}}{2}, a b=\frac{1}{2}$, solving these equations yields $a=1, b=\frac{1}{2}$, so the equation of the ellipse $C$ is $x^{2}+4 y^{2}=1$.
$\cdots \cdots$ (5 points)
(2) Let the equation of the line $l$ be $x=m y+\lambda$, and let the coordinates of the two intersection points be $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$.
From $\overrightarrow{A P}=2 \overrightarrow{P B}$, we get $y_{1}=-2 y_{2}$. $\qquad$
$\qquad$ (10 points)
Solving the system of equations $\left\{\begin{array}{l}x^{2}+4 y^{2}=1 \\ x=m y+\lambda\end{array}\right.$, we obtain $\left(m^{2}+4\right) y^{2}+2 \lambda m y+\lambda^{2}-1=0$ $\qquad$
Clearly, $y_{1}, y_{2}$ are the two distinct real roots of equation (2), so we have
$$
(2 \lambda m)^{2}-4\left(\lambda^{2}-1\right)\left(m^{2}+4\right)>0 \Rightarrow m^{2}>4\left(\lambda^{2}-1\right)
$$
$\qquad$
By Vieta's formulas, we get $y_{1}+y_{2}=-\frac{2 \lambda m}{m^{2}+4}, y_{1} y_{2}=\frac{\lambda^{2}-1}{m^{2}+4}$, combining with (1) we have
$2\left(\frac{2 \lambda m}{m^{2}+4}\right)^{2}+\frac{\lambda^{2}-1}{m^{2}+4}=0 \Rightarrow m^{2}=\frac{4\left(1-\lambda^{2}\right)}{9 \lambda^{2}-1}$
$\cdots \cdots$ (15 points)

Also, $\lambda= \pm 1, \lambda= \pm \frac{1}{3}$ do not satisfy the problem.
Substituting (4) into (3) we get $\frac{4\left(1-\lambda^{2}\right)}{9 \lambda^{2}-1}>4\left(\lambda^{2}-1\right) \Rightarrow \frac{1}{9}<\lambda^{2}<1 \Rightarrow \lambda \in\left(-1,-\frac{1}{3}\right) \cup\left(\frac{1}{3}, 1\right)$. (20 points)