If $\left(1-x^3\right)^{10}=\sum\limits_{\mathrm{r}=0}^{10} \mathrm{a}_{\mathrm{r}} x^{\mathrm{r}}(1-x)^{30-2 \mathrm{r}}$, then $\frac{9 \mathrm{a}_9}{\mathrm{a}_{10}}$ is equal to $\_\_\_\_$ .

Let the line $x-y=4$ intersect the circle $\mathrm{C}:(x-4)^2+(y+3)^2=9$ at the points Q and R . If $\mathrm{P}(\alpha, \beta)$ is a point on C such that $\mathrm{PQ}=\mathrm{PR}$, then $(6 \alpha+8 \beta)^2$ is equal to $\_\_\_\_$ .

Let the image of the point $\mathrm{P}(0,-5,0)$ in the line $\frac{x-1}{2}=\frac{y}{1}=\frac{z+1}{-2}$ be the point R and the image of the point $\mathrm{Q}\left(0, \frac{-1}{2}, 0\right)$ in the line $\frac{x-1}{-1}=\frac{y+9}{4}=\frac{z+1}{1}$ be the point S . Then the square of the area of the parallelogram PQRS is $\_\_\_\_$ .

Let $f(x)=\left\{\begin{array}{ll}x^3+8 ; & x<0, \\ x^2-4 ; & x \geq 0,\end{array}\right.$ and $g(x)= \begin{cases}(x-8)^{1 / 3} ; & x<0, \\ (x+4)^{1 / 2} ; & x \geq 0 .\end{cases}$

Then the number of points, where the function $g \circ f$ is discontinuous, is $\_\_\_\_$ .