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

If $\overrightarrow{\mathrm{a}},\overrightarrow{\mathrm{b}},\overrightarrow{\mathrm{c}}$ are linearly dependent vectors, then $\overrightarrow{\mathrm{c}}$ should be a linear combination of $\overrightarrow{a}\text{ and }\overrightarrow{b}\text{. }$

Let $\overrightarrow{\mathrm{c}}=\mathrm{p}\overrightarrow{\mathrm{a}}+\mathrm{q}\overrightarrow{\mathrm{b}}$

$\text{ i.e. }\widehat{\mathrm{i}}+\alpha \widehat{\mathrm{j}}+\beta \widehat{\mathrm{k}}=\mathrm{p}(\widehat{\mathrm{i}}+\widehat{\mathrm{j}}+\widehat{\mathrm{k}})+\mathrm{q}(4\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}+4\widehat{\mathrm{k}})$

Equating coefficients of $\widehat{\mathrm{i}},\widehat{\mathrm{j}},\widehat{\mathrm{k}}$ we get 

$1=p+4q,\alpha =p+3q,\beta =p+4q$

from first and third, $\beta =1$

Now, $\left|\overrightarrow{c}\right|=\sqrt{3}$

$\begin{array}{l}\therefore 1+{\alpha }^2+{\beta }^2=3\\ \Rightarrow 1+{\alpha }^2+1=3\\ \Rightarrow \alpha =\pm 1\end{array}$

Hence, $\alpha =\pm 1,\beta =1$