If the constant term in the expansion of $$\left(\frac{\sqrt[5]{3}}{x}+\frac{2 x}{\sqrt[3]{5}}\right)^{12}, x \neq 0$$, is $$\alpha \times 2^8 \times \sqrt[5]{3}$$, then $$25 \alpha$$ is equal to :

Let $$\vec{a}=2 \hat{i}+5 \hat{j}-\hat{k}, \vec{b}=2 \hat{i}-2 \hat{j}+2 \hat{k}$$ and $$\vec{c}$$ be three vectors such that $$(\vec{c}+\hat{i}) \times(\vec{a}+\vec{b}+\hat{i})=\vec{a} \times(\vec{c}+\hat{i})$$. If $$\vec{a} \cdot \vec{c}=-29$$, then $$\vec{c} \cdot(-2 \hat{i}+\hat{j}+\hat{k})$$ is equal to:

Let $$\mathrm{A}(-1,1)$$ and $$\mathrm{B}(2,3)$$ be two points and $$\mathrm{P}$$ be a variable point above the line $$\mathrm{AB}$$ such that the area of $$\triangle \mathrm{PAB}$$ is 10. If the locus of $$\mathrm{P}$$ is $$\mathrm{a} x+\mathrm{by}=15$$, then $$5 \mathrm{a}+2 \mathrm{~b}$$ is :

The values of $$m, n$$, for which the system of equations

$$\begin{aligned} & x+y+z=4, \\ & 2 x+5 y+5 z=17, \\ & x+2 y+\mathrm{m} z=\mathrm{n} \end{aligned}$$

has infinitely many solutions, satisfy the equation :