Suppose $A$ and $B$ are the coefficients of $30^{\text {th }}$ and $12^{\text {th }}$ terms respectively in the binomial expansion of $(1+x)^{2 \mathrm{n}-1}$. If $2 \mathrm{~A}=5 \mathrm{~B}$, then n is equal to:

Let the position vectors of three vertices of a triangle be $4 \vec{p}+\vec{q}-3 \vec{r},-5 \vec{p}+\vec{q}+2 \vec{r}$ and $2 \vec{p}-\vec{q}+2 \vec{r}$. If the position vectors of the orthocenter and the circumcenter of the triangle are $\frac{\vec{p}+\vec{q}+\vec{r}}{4}$ and $\alpha \vec{p}+\beta \vec{q}+\gamma \vec{r}$ respectively, then $\alpha+2 \beta+5 \gamma$ is equal to :

If $7=5+\frac{1}{7}(5+\alpha)+\frac{1}{7^2}(5+2 \alpha)+\frac{1}{7^3}(5+3 \alpha)+\ldots \ldots \ldots \ldots \infty$, then the value of $\alpha$ is :

If the equation of the parabola with vertex $\mathrm{V}\left(\frac{3}{2}, 3\right)$ and the directrix $x+2 y=0$ is $\alpha x^2+\beta y^2-\gamma x y-30 x-60 y+225=0$, then $\alpha+\beta+\gamma$ is equal to :