(11) If the line $3 x \sin ^{2} \alpha+y \cos ^{2} \alpha-3=0$ intersects the hyperbola $x^{2}-y^{2}=1$ at only one point, then the coordinates of this point are $\qquad$ .
$11(1,0)$ or $\left(\frac{17}{8}, \frac{15}{8}\right)$ Hint: (1) When $\cos ^{2} \alpha=0$, $\sin ^{2} \alpha=1$, the equation of the line is $x=1$. Substituting into $x^{2}-y^{2}=1$ yields the unique intersection point $(1,0)$ of the line and the hyperbola at this time;
(2) When $\cos ^{2} \alpha \neq 0$, let the equation of the line be $y=k(x-1)+3$, where $k=-3 \tan ^{2} \alpha \leqslant 0$.
(i) When $k=-1$, the equation of the line is $y=4-x$. Substituting into $x^{2}-y^{2}=1$ yields the only common point $\left(\frac{17}{8}, \frac{15}{8}\right)$ of the line and the hyperbola at this time.
(ii) When $k \neq-1$, then from $\left\{\begin{array}{l}y=k(x-1)+3, \\ x^{2}-y^{2}=1,\end{array}\right.$ we get
$$
\begin{array}{l}
\left(1-k^{2}\right) x^{2}+\left(2 k^{2}-6 k\right) x-k^{2}+6 k-10=0, \\
\begin{aligned}
\Delta & =\left(2 k^{2}-6 k\right)^{2}-4\left(1-k^{2}\right)\left(-k^{2}+6 k-10\right) \\
& =-24 k+40,
\end{aligned}
\end{array}
$$

Solving $\Delta=0$ yields $k=\frac{5}{3}>0$, which does not meet the requirement.
In summary, when the line and the hyperbola have only one common point, the coordinates of this common point are either $(1,0)$ or $\left(\frac{17}{8}, \frac{15}{8}\right)$.