Given: $\overrightarrow a, \:\overrightarrow b\:\;and\:\:\overrightarrow c$ are linearly dependent,
$\Rightarrow\:\left |\begin {array}{ccc}1 & 1 & 1\\ 4 & 3 & 4\\1 & \alpha & \beta\end {array}\right|=0$
$\Rightarrow\:3\beta-4\alpha-4\beta+4+4\alpha-3=0$
$\Rightarrow\:1-\beta=0$ or $\beta=1$
Also given that $|\overrightarrow c|=\sqrt 3$
$\Rightarrow\:1+\alpha^2+\beta^2=3$
$\Rightarrow\:\alpha=\pm 1$
$\therefore\:(\alpha,\beta)=(\pm 1,1)$