Since, $\alpha$ is a root of
$\therefore$
$$
\begin{array}{l}
x^{2}+3 p^{2} x+5 q^{2}=0 \\
\alpha^{2}+3 p^{2} \alpha+5 q^{2}=0
\end{array}
$$
and $\beta$ is a root of
$\therefore$
$$
\begin{array}{l}
x^{2}+9 p^{2} x+15 q^{2}=0 \\
\beta^{2}+9 p^{2} \beta+15 q^{2}=0
\end{array}
$$
Let $f(x)=x^{2}+6 p^{2} x+10 q^{2}$
$\begin{aligned} \text { Then, } f(\alpha) &=\alpha^{2}+6 p^{2} \alpha+10 q^{2} \\ &=\left(\alpha^{2}+3 p^{2} \alpha+5 q^{2}\right)+3 p^{2} \alpha+5 q^{2} \end{aligned}$
$=0+3 p^{2} \alpha+5 q^{2}$
$\Rightarrow \quad f(\alpha)>0$
and $\quad f(\beta)=\beta^{2}+6 p^{2} \beta+10 q^{2}$
$=\left(\beta^{2}+9 p^{2} \beta+15 q^{2}\right)-\left(3 p^{2} \beta+5 q^{2}\right)$
$=0-\left(3 p^{2} \beta+5 q^{2}\right) \quad$ I from Eq. (ii)]
$\Rightarrow \quad f(\beta) < 0$
Thus, $f(x)$ is a polynomial such that $f(\alpha)>0$ and $f(\beta) < 0$
Therefore, there exists $\gamma$ satisfying $\alpha < \gamma < \beta$ such that $f(\gamma)=0$