Given, $f(x)=2 x^{3}-9 a x^{2}+12 a^{2} x+1$ attains maximum and minimum at $p$ and $q$ respectively. $\therefore$
$$
f^{\prime}(p)=0, f^{\prime}(q)=0
$$
Now,
$$
\begin{array}{c}
f^{\prime \prime}(p) < 0 \text { and } f^{\prime \prime}(q)>0 \\
f^{\prime}(p)=0
\end{array}
$$
and
$$
\begin{array}{r}
f^{\prime}(q)=0 \\
6 p^{2}-18 a p+12 a^{2}=0 \\
6 q^{2}-18 a q+12 a^{2}=0
\end{array}
$$
and
$$
\begin{array}{ll}
\Rightarrow & p^{2}-3 a p+2 a^{2}=0 \\
\text { and } & q^{2}-3 a q+2 a^{2}=0
\end{array}
$$
$p=a, 2 a, q=a, 2 a$
$\begin{array}{lrl}\Rightarrow & p=a, 2 a, q=a, \\ \text { Now, } & f^{\prime \prime}(p) < 0\end{array}$
$\Rightarrow \quad 12 p-18 a < 0$
$\Rightarrow \quad p < \frac{3}{2} a$
and $f^{\prime \prime}(q)>0 \Rightarrow 12 q-18 a>0$
$\Rightarrow \quad q>\frac{3}{2} a$
From Eqs. (i), (ii) and (iii), we get $\begin{aligned} p &=a, q=2 a \\ \text { Now, } \quad p^{2} &=q \end{aligned}$
$$
\begin{array}{lc}
\Rightarrow & a^{2}=2 a \\
\Rightarrow & a=0,2
\end{array}
$$
But for $a=0, f(x)=2 x^{3}+1$ which does not
attains a maximum or minimum for any value of $x$. Hence, $a=2$.
$$