For the reaction, XA +YB $\to$ ZC, $\text{ if }\frac{-\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{d}t}=\frac{-\mathrm{d}\left[\mathrm{B}\right]}{\mathrm{dt}}=\frac{1.5\mathrm{d}\left[\mathrm{C}\right]}{\mathrm{dt}}$, then the correct  statement among the following is :

$\begin{array}{l}\mathrm{XA}+\mathrm{YB}\stackrel{ }{\to }\mathrm{ZC}\\ \frac{-1}{\mathrm{X}}\frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{dt}}=\frac{-1}{\mathrm{Y}}\frac{\mathrm{d}\left[\mathrm{B}\right]}{\mathrm{dt}}=\frac{+1}{\mathrm{Z}}\frac{\mathrm{d}\left[\mathrm{C}\right]}{\mathrm{dt}}\\ \mathrm{Given},\frac{-\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{dt}}=\frac{-\mathrm{d}\left[\mathrm{B}\right]}{\mathrm{dt}}=1.5\frac{\mathrm{d}\left[\mathrm{C}\right]}{\mathrm{dt}}\\ \frac{-\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{dt}}=\frac{-\mathrm{d}\left[\mathrm{B}\right]}{\mathrm{dt}}=\frac{3}{2}\frac{\mathrm{d}\left[\mathrm{C}\right]}{\mathrm{dt}}\end{array}$

Dividing throughout  by factor three

$-\frac{1}{3}\frac{\mathrm{d}\left[\mathrm{A}\right]}{\mathrm{dt}}=-\frac{1}{3}\frac{\mathrm{d}\left[\mathrm{B}\right]}{\mathrm{dt}}=+\frac{1}{2}\frac{\mathrm{d}\left[\mathrm{C}\right]}{\mathrm{dt}}$

Thus the stoichiometric reaction would be 

$3\mathrm{A}+3\mathrm{B}\stackrel{ }{\to }2\mathrm{C}$

X = Y = 3 ; Z = 2