How do you simplify ${{\left( 3a{{b}^{2}}c \right)}^{2}}{{\left( -2{{a}^{2}}{{b}^{4}} \right)}^{2}}{{\left( {{a}^{4}}{{c}^{2}} \right)}^{3}}{{\left( {{a}^{2}}{{b}^{4}}{{c}^{5}} \right)}^{2}}{{\left( 2{{a}^{3}}{{b}^{2}}{{c}^{4}} \right)}^{3}}$?

Hint: We first explain the process of exponents and indices. We find the general form. Then we explain the different binary operations on exponents. We use the identities to find the simplified form of the numbers with positive exponents.

Complete step-by-step solution:
We know the exponent form of the number $a$ with the exponent being $n$ can be expressed as ${{a}^{n}}$. In case the value of $n$ becomes negative, the value of the exponent takes its inverse value. The formula to express the form is ${{a}^{-n}}=\dfrac{1}{{{a}^{n}}},n\in {{\mathbb{R}}^{+}}$.
If we take two exponential expressions where the exponents are $m$ and $n$.
Let the numbers be ${{a}^{m}}$ and ${{a}^{n}}$. We take multiplication of these numbers.
The indices get added. So, ${{a}^{m+n}}={{a}^{m}}\times {{a}^{n}}$.
The division works in an almost similar way. The indices get subtracted. So, $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$.
We also have the identity of ${{\left( ab \right)}^{n}}={{a}^{n}}{{b}^{n}}$.
We have been given a bunch of numbers and their indices.
We solve these one by one.
We get the first term where ${{\left( 3a{{b}^{2}}c \right)}^{2}}={{3}^{2}}{{a}^{2}}{{\left( {{b}^{2}} \right)}^{2}}{{c}^{2}}=9{{a}^{2}}{{b}^{4}}{{c}^{2}}$
Then we have ${{\left( -2{{a}^{2}}{{b}^{4}} \right)}^{2}}={{\left( -2 \right)}^{2}}{{\left( {{a}^{2}} \right)}^{2}}{{\left( {{b}^{4}} \right)}^{2}}=4{{a}^{4}}{{b}^{8}}$
Similarly,
\[\begin{align}
  & {{\left( {{a}^{4}}{{c}^{2}} \right)}^{3}}={{\left( {{a}^{4}} \right)}^{3}}{{\left( {{c}^{2}} \right)}^{3}}={{a}^{12}}{{c}^{6}} \\
 & {{\left( {{a}^{2}}{{b}^{4}}{{c}^{5}} \right)}^{2}}={{\left( {{a}^{2}} \right)}^{2}}{{\left( {{b}^{4}} \right)}^{2}}{{\left( {{c}^{5}} \right)}^{2}}={{a}^{4}}{{b}^{8}}{{c}^{10}} \\
 & {{\left( 2{{a}^{3}}{{b}^{2}}{{c}^{4}} \right)}^{3}}={{2}^{3}}{{\left( {{a}^{3}} \right)}^{3}}{{\left( {{b}^{2}} \right)}^{3}}{{\left( {{c}^{4}} \right)}^{3}}=8{{a}^{9}}{{b}^{6}}{{c}^{12}} \\
\end{align}\]
Now all these numbers are in multiplied form. We take the same number together. Their powers get added.
$\begin{align}
  & {{\left( 3a{{b}^{2}}c \right)}^{2}}{{\left( -2{{a}^{2}}{{b}^{4}} \right)}^{2}}{{\left( {{a}^{4}}{{c}^{2}} \right)}^{3}}{{\left( {{a}^{2}}{{b}^{4}}{{c}^{5}} \right)}^{2}}{{\left( 2{{a}^{3}}{{b}^{2}}{{c}^{4}} \right)}^{3}} \\
 & =9{{a}^{2}}{{b}^{4}}{{c}^{2}}\times 4{{a}^{4}}{{b}^{8}}\times {{a}^{12}}{{c}^{6}}\times {{a}^{4}}{{b}^{8}}{{c}^{10}}\times 8{{a}^{9}}{{b}^{6}}{{c}^{12}} \\
 & =\left( 9\times 4\times 8 \right)\times {{a}^{2+4+12+4+9}}\times {{b}^{4+8+8+6}}\times {{c}^{2+6+10+12}} \\
 & =288{{a}^{31}}{{b}^{26}}{{c}^{30}} \\
\end{align}$
Therefore, simplified form of ${{\left( 3a{{b}^{2}}c \right)}^{2}}{{\left( -2{{a}^{2}}{{b}^{4}} \right)}^{2}}{{\left( {{a}^{4}}{{c}^{2}} \right)}^{3}}{{\left( {{a}^{2}}{{b}^{4}}{{c}^{5}} \right)}^{2}}{{\left( 2{{a}^{3}}{{b}^{2}}{{c}^{4}} \right)}^{3}}$ is $288{{a}^{31}}{{b}^{26}}{{c}^{30}}$.

Note: The addition and subtraction for exponents works for taking common terms out depending on the values of the indices.
For numbers ${{a}^{m}}$ and ${{a}^{n}}$, we have ${{a}^{m}}\pm {{a}^{n}}={{a}^{m}}\left( 1\pm {{a}^{n-m}} \right)$.the relation is independent of the values of $m$ and $n$. We need to remember that the condition for ${{a}^{m}}={{a}^{n}}\Rightarrow m=n$ is that the value of $a\ne 0,\pm 1$.