If $\overrightarrow{a}=\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k},$ $\overrightarrow{b}=4\overrightarrow{i}+3\overrightarrow{j}+4\overrightarrow{k}$, $\overrightarrow{c}=\overrightarrow{i}+\alpha \overrightarrow{j}+\beta \overrightarrow{k}$are linearly dependent vectors and $\left|\overrightarrow{c}\right|=\sqrt{3}$ then

$\left|\overrightarrow{C}\right|=\sqrt{3}\Rightarrow {\left|\overrightarrow{C}\right|}^2=3$

$1+{\alpha }^2+{\beta }^2=3\Rightarrow {\alpha }^2+{\beta }^2=2.......\left(1\right)$

Given vectors are L.D

$\therefore \left|\begin{array}{ccc}1& 1& 1\\ 4& 3& 4\\ 1& \alpha & \beta \end{array}\right|=0$

Det by $R_3\to$

$1\left(1\right)-\alpha \left(0\right)+\beta (-1)=0$

$\therefore \beta =1$ From (1) ${\alpha }^2+1=2\Rightarrow \alpha =\pm 1$