Simplify the given expression $\dfrac{{6{x^2}{y^{12}}{z^6}}}{{{{\left( {2{x^2}yz} \right)}^3}}}$ for all $xyz \ne 0$.
$
  (a){\text{ }}\dfrac{{{y^4}{z^2}}}{{{x^3}}} \\
  (b){\text{ }}\dfrac{{{y^9}{z^4}}}{{{x^4}}} \\
  (c){\text{ }}\dfrac{{{y^9}{z^3}}}{{2{x^3}}} \\
  (d){\text{ }}\dfrac{{3{y^4}{z^2}}}{{4{x^3}}} \\
  (e){\text{ }}\dfrac{{3{y^9}{z^3}}}{{4{x^4}}} \\
 $

Hint: Start with the simplification of the denominator part and take the terms common in both numerator and denominator, this will help simplify the given expression.

Complete Step-by-Step solution:
Given linear equation is $\dfrac{{6{x^2}{y^{12}}{z^6}}}{{{{\left( {2{x^2}yz} \right)}^3}}}$ for all $xyz \ne 0$
Now we have to simplify this so we first simplify the denominator of given equation,
$ \Rightarrow \dfrac{{6{x^2}{y^{12}}{z^6}}}{{{{\left( {2{x^2}yz} \right)}^3}}} = \dfrac{{6{x^2}{y^{12}}{z^6}}}{{8{x^6}{y^3}{z^3}}}$
Now take $2{x^2}{y^3}{z^3}$ common from numerator and denominator we have,
$ \Rightarrow \dfrac{{6{x^2}{y^{12}}{z^6}}}{{8{x^6}{y^3}{z^3}}} = \dfrac{{\left( {2{x^2}{y^3}{z^3}} \right)3{y^9}{z^3}}}{{\left( {2{x^2}{y^3}{z^3}} \right)4{x^4}}}$
Now cancel out the common terms from numerator and denominator we have,
$ \Rightarrow \dfrac{{\left( {2{x^2}{y^3}{z^3}} \right)3{y^9}{z^3}}}{{\left( {2{x^2}{y^3}{z^3}} \right)4{x^4}}} = \dfrac{{3{y^9}{z^3}}}{{4{x^4}}}$
So this is the required answer.
Hence option (E) is correct.

Note: There are some properties of power evaluation that are very crucial in solving problems of this kind, some of them are, powers with exponent one that is any value raised to 1 results in the same value. In case of other power that is high power to another power, the base is maintained same and the powers in the exponents are multiplied. There are many other properties. It is advised to go through them all.