The sum of all oxidation values must be equal to zero. Let, number of atoms of X, Y and Z are respectively a, b and c.

Then, $(+3)\text{a}+(+5)\text{b}+(-2)\text{c}=0$

Or

$3\text{a}+5\text{b}-2\text{c}=0$

Now, putting values of a, b and c for all the given options, which will give sum equal to zero, will be our formula.

(a) For, ${\text{X}}_2\text{Y}\text{Z}$

$\text{a}=2$

$\text{b}=1$

$\text{c}=1$

Thus, $3\text{a}+5\text{b}-2\text{c}=0$

$3\times 2+5-2\ne 0$

(b) For, $\text{X}\text{Y}{\text{Z}}_3$

$\text{a}=1,\text{b}=1,\text{c}=3$

$3\text{a}+5\text{b}-2\text{c}=0$

$3+5-6\ne 0$

(c) ${\text{X}}_3{\left(\text{Y}{\text{Z}}_4\right)}_3$

$\text{a}=3,\text{b}=3,\text{c}=12$

$3\text{a}+5\text{b}-2\text{c}=0$

$(3\times 3+5\times 3)-(2\times 12)=0$

$9+15-24$

$24-24=0$

(d) ${\text{X}}_2{\text{Y}}_2\text{Z}$

$\text{a}=2,\text{b}=2,\text{c}=1$

$3\text{a}+5\text{b}-2\text{c}=0$

$(3\times 2)+(5\times 2)-(2\times 1)=0$

$6+10-2\ne 0$

Hence, option (c) is the correct answer.