Simplify sqrt(12) + sqrt(3), determine if the result is rational or irrational, and find two consecutive integers it lies between.

Using the approximate value $\sqrt{3} \approx 1.73205$:

Rounding to two decimal places:

A number is rational if it can be expressed as a fraction $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$. A number is irrational if it cannot be expressed in this form.
Since $\sqrt{3}$ is an irrational number, and the product of a non-zero rational number (3) and an irrational number ($\sqrt{3}$) is always irrational, $A$ is an irrational number.

The value of A is irrational.
Justification: $\sqrt{3}$ is an irrational number, and the product of a non-zero rational number (3) and an irrational number ($\sqrt{3}$) is always irrational.

We know that $5^2 = 25$ and $6^2 = 36$.
So, $25 < 27 < 36$.

Since $A = \sqrt{27}$, the value of A lies between 5 and 6.

The value of A lies between the consecutive integers $\boxed{5 \text{ and } 6}$.