Given

$\rm 3^{x+2} + 3^{x+1} = 1 \frac{1}{3}$

By using the law of indices, $\rm a^{m+n} = a^m \cdot a^n$, we get,

$\rm or, 3^x \cdot 3^2 + 3^x \cdot 3^1 = \frac{3 \cdot 1 + 1}{3}$

$\rm or, 3^x ( 3^2 + 3^1) = \frac{4}{3}$

$\rm or, 3^x (9 + 3) = \frac{4}{3}$

$\rm or, 3^x (12) = \frac{4}{3}$

Dividing both sides of the equation by 12, we get,

$\rm or, 3^x \frac{12}{12} = \frac{4}{3} \cdot \frac{1}{12}$

$\rm or, 3^x = \frac{1}{9}$

$\rm or, 3^x = \frac{1}{3^2}$

By the law of indices, $\rm a^{-m} = \frac{1}{a^m}$, we get,

$\rm or 3^x = 3^{-2}$

The bases of the terms on both sides of the equation are the same, so we equate their powers.

$\rm \therefore x = -2$

Hence, the required value of x is -2.