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

Solution :
$$\eqalign{ & {\text{Here, }}\overrightarrow c = t\overrightarrow a + s\overrightarrow b {\text{ and }}\left| {\overrightarrow c } \right| = \sqrt 3 = \sqrt {1 + {\alpha ^2} + {\beta ^2}} \cr & {\text{Now, }}\overrightarrow i + \alpha \overrightarrow j + \beta \overrightarrow k = t\left( {\overrightarrow i + \overrightarrow j + \overrightarrow k } \right) + s\left( {4\overrightarrow i + 3\overrightarrow j + 4\overrightarrow k } \right) \cr & \Rightarrow 1 = t + 4s,\,\,\alpha = t + 3s,\,\,\beta = t + 4s \cr & \therefore {\text{ we get }}3 = 1 + {\alpha ^2} + {\beta ^2}\,\, \Rightarrow 2 = {\left( {t + 3s} \right)^2} + {\left( {t + 4s} \right)^2} \cr & {\text{or }}2 = {\left( {1 - 4s + 3s} \right)^2} + {\left( {1 - 4s + 4s} \right)^2} \cr & {\text{or }}1 = {\left( {1 - s} \right)^2} \cr & \therefore \,1 - s = \pm 1\,\,\,\therefore \,s = 1 \mp 1 = 0,\,2 \cr & \therefore \,t = 1 - 4s = 1,\, - 7 \cr & \therefore \,\,\alpha = t + 3s = 1 + 3.0,\, - 7 + 3.2 = 1,\, - 1\,\,{\text{and}} \cr & \,\,\,\,\,\,\,\,\beta = t + 4s = 1 + 4.0,\, - 7 + 4.2 = 1,\,1 \cr} $$