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

Given that $\overrightarrow{\mathrm{a}} =\hat{i} + \hat{j}+ \hat{k}ˆ,\overrightarrow{\mathrm{b}} =4\hat{i} +3 \hat{j}+4 \hat{k}$ and  $\overrightarrow{c} = \hat{i} +\alpha \hat{j}+\beta \hat{k}$ are linearly dependent.

$\left|\begin{array}{cccc}1& 1& 1& \\ 4& 3& 4& \\ 1& \alpha & \beta & \end{array}\right|=0$ Or $1-\beta =0$ Or $\beta =1$

Also given that  $\left|\overrightarrow{c}\right| = \sqrt{3} \Rightarrow 1 + {\alpha }^2 + {\beta }^2 = 3$ substituting the value of  $\beta$, we get  $a^2=1 or a=+1$