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**Hint:**Here, we will use the given equation, to find the relationship between \[x\], \[y\] and \[z\]. Then, we use the obtained relation and solve each equation. We will then check the answer of each option whether they meet the condition given in the question. The option that satisfies the condition is the required answer.

**Complete step-by-step answer:**It is given that \[{x^x} \cdot {y^y} \cdot {z^z} = {x^y} \cdot {y^z} \cdot {z^x}\].

Hence, comparing the powers of \[x\], \[y\] and \[z\] respectively on LHS and RHS, we observe that,

\[x = y\], \[y = z\] and \[z = x\]

This means that \[x = y = z\].

Now, we have to find that which of the given options cannot be true for any of the possible value of \[x\], \[y\] and \[z\], where we know that \[x\], \[y\] and \[z\] are positive integers greater than 1.

A. \[xyz = 27\]

Now, we know that \[x = y = z\].

Hence, substituting this in \[xyz = 27\], we get

\[ \Rightarrow x \cdot x \cdot x = 27\]

\[ \Rightarrow {x^3} = 27\]

Taking cube root on both sides, we get

\[ \Rightarrow x = 3\]

$\therefore x=y=z=3$

Hence, this option is true for any of the possible values of \[x\], \[y\] and \[z\], where we know that \[x\], \[y\] and \[z\] are positive integers greater than 1.

B. \[xyz = 1728\]

Now, we know that \[x = y = z\]

Hence, substituting this in \[xyz = 1728\], we get

\[ \Rightarrow x \cdot x \cdot x = 1728\]

\[ \Rightarrow {x^3} = 1728\]

Taking cube root on both sides, we get

\[ \Rightarrow x = 12\]

$\therefore x=y=z=12$

Hence, this option is true for any of the possible values of \[x\], \[y\]and \[z\], where we know that \[x\], \[y\]and \[z\]are positive integers greater than 1.

C. \[x + y + z = 32\]

Now, we know that \[x = y = z\]

Hence, substituting this in \[x + y + z = 32\], we get

\[ \Rightarrow x + x + x = 32\]

\[ \Rightarrow 3x = 32\]

Dividing both sides by 3, we get

\[ \Rightarrow x = 10.667\]

$\therefore x=y=z=10.667$

Hence, this option cannot be true for any of the possible values of \[x\], \[y\] and \[z\], as we know that \[x\], \[y\] and \[z\]are positive integers greater than 1.

Because in this case, the values of \[x\], \[y\] and \[z\]are not integers.

Hence, this is the required answer.

But, we will check for the last part as well.

D. \[x + y + z = 12\]

Now, we know that \[x = y = z\]

Hence, substituting this in \[x + y + z = 12\], we get

\[ \Rightarrow x + x + x = 12\]

\[ \Rightarrow 3x = 12\]

Dividing both sides by 3, we get

\[ \Rightarrow x = 4\]

$\therefore x=y=z=4$

Hence, this option is also true for any of the possible values of \[x\], \[y\] and \[z\], as we know that \[x\], \[y\] and \[z\] are positive integers greater than 1.

Therefore, from the given options only option C cannot be true as it does not give an integer value of \[x\], \[y\] and \[z\].

**Hence, Option C is the required answer.**

**Note:**Integers are those numbers which are either positive i.e. starting from 1 to infinity, or negative, i.e. from \[ - 1\] to \[ - \infty \] or its value is 0. Integers cannot be written in the form of fractions or decimals. Hence, this means that they are represented in the form of a number as a whole. Here, in all options, we got numbers greater than 1 but they are decimal numbers except for the number obtained in option C. So, we had chosen option C as the answer because it satisfied the given condition.

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