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 :

<p>$$\begin{aligned} & \left(\frac{\sqrt[5]{3}}{x}+\frac{2 x}{\sqrt[5]{3}}\right)^{12} \\ & T_{r+1}={ }^{12} C r\left(\frac{\sqrt[5]{3}}{x}\right)^{12-r}\left(\frac{2 x}{\sqrt[3]{5}}\right)^r \end{aligned}$$</p> <p>For constant term $-12+r+r=0$</p> <p>$$\begin{aligned} & \Rightarrow \quad r=6 \\ & \therefore \quad \text { Constant term }={ }^{12} C_6 \frac{(3)^{\frac{6}{5}}}{(5)^{\frac{6}{3}}}(2)^6 \end{aligned}$$</p> <p>$$\begin{aligned} & ={ }^{12} C_6 \times \frac{2^6}{25} \times 3.3^{\frac{1}{5}} \\ & =\frac{231}{25} \times 2^8 \cdot 3^{\frac{1}{5}} \cdot 3 \\ & =\frac{693}{25} \cdot 2^8 \sqrt[5]{3} \\ & \therefore \quad \alpha=\frac{693}{25} \\ & 25 \alpha=693 \end{aligned}$$</p>