1. (16 points) Does there exist a real number $a$, such that the line $y=a x+1$ intersects the hyperbola $3 x^{2}-y^{2}=1$ at two points $A$ and $B$, and the circle with diameter $AB$ passes exactly through the origin of the coordinate system?
1. Let points $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$.
Substitute $y=a x+1$ into $3 x^{2}-y^{2}=1$, and eliminate $y$ to get
$$
\left(3-a^{2}\right) x^{2}-2 a x-2=0 \text {. }
$$

Then $x_{1}$ and $x_{2}$ are the roots of the equation
$$
\left(3-a^{2}\right) x^{2}-2 a x-2=0
$$

By Vieta's formulas, we have
$$
x_{1}+x_{2}=\frac{2 a}{3-a^{2}}, x_{1} x_{2}=-\frac{2}{3-a^{2}} \text {. }
$$

The circle with diameter $AB$ passes through the origin $O$
$$
\begin{array}{l}
\Leftrightarrow \overrightarrow{O A} \cdot \overrightarrow{O B}=0 \\
\Leftrightarrow x_{1} x_{2}+y_{1} y_{2}=0 \\
\Leftrightarrow x_{1} x_{2}+\left(a x_{1}+1\right)\left(a x_{2}+1\right)=0 .
\end{array}
$$

Substitute equation (1) and simplify to get
$$
\frac{1-a^{2}}{3-a^{2}}=0 \Rightarrow a= \pm 1 \text {. }
$$

Upon verification, $a= \pm 1$ satisfies the conditions of the problem.