Calculating Mean $=\frac{3+7+12+a+43-a}{5}=13$

Variance $=\frac{9+49+144+a^2+\left(43-a\right){}^2}{5}-{13}^2\in N$

$\Rightarrow \frac{2a^2-a+1}{5}=n$ (where $n\in N$)

$\Rightarrow 2a^2-a+1-5n=0$ must have solution as natural numbers

So, $D=40n-7$ always has $3$ at unit place for any value of $n$

$\Rightarrow D$ can't be perfect square

So, $a$ can't be integer.