Solve the following equation \[{{x}^{3}}{{y}^{2}}z=12,{{x}^{3}}y{{z}^{3}}=54,{{x}^{7}}{{y}^{3}}{{z}^{2}}=72\].
A. x = 1, y = 2, z = 3
B. x = 1, y = 1, z = 2
C. x = 1, y = 2, z= 3
D. x = -1, y = 2, z = 4

Hint: In this question, we will find the relation among x, y and z using the given equation. We will divide one equation by another to get a relation among x, y and z.

Complete step-by-step answer:

We have been given the equation as follows:
\[{{x}^{3}}{{y}^{2}}z=12.....(1)\]
\[{{x}^{3}}y{{z}^{3}}=54....(2)\]
\[{{x}^{7}}{{y}^{3}}{{z}^{2}}=72.....(3)\]
Now we will divide equation (1) by equation (2).
We will get as follows:
\[\begin{align}
  & \dfrac{{{x}^{3}}{{y}^{2}}z}{{{x}^{3}}y{{z}^{3}}}=\dfrac{12}{54} \\
 & \dfrac{y}{{{z}^{2}}}=\dfrac{2}{9}.....(4) \\
\end{align}\]
Once again, we will square both sides of the equation (1) and divide it by equation (3).
We will get as follows:
\[\begin{align}
  & \dfrac{{{\left( {{x}^{3}}{{y}^{2}}z \right)}^{2}}}{{{x}^{7}}{{y}^{3}}{{z}^{2}}}=\dfrac{{{\left( 12 \right)}^{2}}}{72} \\
 & \dfrac{{{x}^{6}}{{y}^{4}}{{z}^{2}}}{{{x}^{7}}{{y}^{3}}{{z}^{2}}}=\dfrac{12\times 12}{72} \\
 & \dfrac{y}{x}=2 \\
 & y=2x......(5) \\
\end{align}\]
Once again we will cube both the sides of the equation (2) and divide it by equation (3).
We will get as follows:
\[\begin{align}
  & \dfrac{{{\left( {{x}^{3}}y{{z}^{3}} \right)}^{3}}}{{{x}^{7}}{{y}^{3}}{{z}^{2}}}=\dfrac{{{\left( 54 \right)}^{3}}}{72} \\
 & \dfrac{{{x}^{9}}{{y}^{3}}{{z}^{9}}}{{{x}^{7}}{{y}^{3}}{{z}^{2}}}=\dfrac{{{\left( 54 \right)}^{3}}}{72} \\
 & {{x}^{2}}{{z}^{7}}=\dfrac{{{\left( 54 \right)}^{3}}}{72}.....(6) \\
\end{align}\]
By using equation (6), we will try to substitute the value of x.
Then we will get as follows:
\[\begin{align}
  & {{\left( \dfrac{y}{2} \right)}^{2}}{{z}^{7}}=\dfrac{{{\left( 54 \right)}^{3}}}{72} \\
 & {{y}^{2}}{{z}^{7}}=\dfrac{{{\left( 54 \right)}^{3}}}{72}\times 4 \\
 & {{y}^{2}}{{z}^{7}}=\dfrac{{{\left( 54 \right)}^{3}}}{18} \\
\end{align}\]
Now by using equation (4) and substituting the value of y in the above equation, we will get as follows:
\[\begin{align}
  &{ {\left( \dfrac{2}{9}{{z}^{2}} \right)}^{2}}{{z}^{7}}=\dfrac{{{\left( 54 \right)}^{3}}}{18} \\
 & \dfrac{{{2}^{2}}}{{{9}^{2}}}\times {{z}^{4}}\times {{z}^{7}}=\dfrac{{{\left( 54 \right)}^{3}}}{18} \\
 & {{z}^{11}}=\dfrac{{{\left( 54 \right)}^{3}}}{18}\times \dfrac{{{9}^{2}}}{{{2}^{2}}} \\
\end{align}\]
\[\begin{align}
  & {{z}^{11}}=\dfrac{54\times 54\times 54\times 9\times 9}{18\times 2\times 2} \\
 & {{z}^{11}}=27\times 27\times 27\times 9 \\
 & {{z}^{11}}={{3}^{3}}\times {{3}^{3}}\times {{3}^{3}}\times {{3}^{2}} \\
 & {{z}^{11}}={{3}^{11}} \\
\end{align}\]
On taking 11th root to both the sides of the equality of the equation, we will get as follows:
\[\begin{align}
  & {{\left( {{z}^{11}} \right)}^{{}^{1}/{}_{11}}}={{\left( {{3}^{11}} \right)}^{{}^{1}/{}_{11}}} \\
 & \therefore z=3 \\
\end{align}\]
Now by substituting the value of z = 3 in the equation (4), we will get as follows:
\[\begin{align}
  & \dfrac{y}{{{z}^{2}}}=\dfrac{2}{9} \\
 & \dfrac{y}{{{\left( 3 \right)}^{2}}}=\dfrac{2}{9} \\
 & y=\dfrac{2}{9}\times 9 \\
 & \therefore y=2 \\
\end{align}\]
Now by substituting the value of y = 2 in the equation (5), we will get as follows:
\[\begin{align}
  & y=2x \\
 & 2=2x \\
 & \therefore x=1 \\
\end{align}\]
Hence we get x = 1, y = 2 and z = 3.
Therefore the correct answer of the above question is option C.

Note: Be careful when we do cube or square of the term as there is a chance that we might forget to square or cube the term properly. Also, we can obtain the answer of the given question by substituting the values of x, y and z from the given options one by one. This can be done only in the competitive examination to save our time.