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 $\qquad$ .
$=11 \cdot(1,0),\left(\frac{17}{8}, \frac{15}{8}\right)$.
For any $\alpha \in \mathbf{R}$, the line always passes through the fixed point $(1,3)$.
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 that the line intersects the hyperbola at only one point $(1,0)$;
When $\cos ^{2} \alpha \neq 0$, let the equation of the line be
$$
y=k(x-1)+3\left(k=-3 \tan ^{2} \alpha \leqslant 0\right) \text {. }
$$
(1) When $k=-1$, the equation of the line is $y=4-x$, substituting into $x^{2}-y^{2}=1$ yields that the line intersects the hyperbola at only one common point $\left(\frac{17}{8}, \frac{15}{8}\right)$.
(2) When $k \neq-1$, from $\left\{\begin{array}{l}y=k(x-1)+3 \text {, } \\ x^{2}-y^{2}=1,\end{array}\right.$ we get $\left(1-k^{2}\right) x^{2}+\left(2 k^{2}-6 k\right) x-k^{2}+6 k-10=0$.
Then $\Delta=24 k-40$.
Solving $\Delta=0$ yields $k=\frac{5}{3}>0$, which does not meet the requirement.
In summary, when the line intersects the hyperbola at only one common point, the common point is $(1,0)$ or $\left(\frac{17}{8}, \frac{15}{8}\right)$.