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Let \overrightarrow{a} = \widehat{1} + \alpha \widehat{j} + 3\widehat{k}  and \overrightarrow{b }= 3\widehat{i} - \alpha \widehat{j} + \widehat{k}. If the area of the parallelogram whose adjacent sides are represented by the vectors \overrightarrow{a} and \overrightarrow{b} is 8 \sqrt{3}  square units, then\overrightarrow{a}.\overrightarrow{b} is equal to ______:
Option: 1 2
Option: 2 1
Option: 3 6
Option: 4 8

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\\\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\alpha \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\ \overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}-\alpha \hat{\mathrm{j}}+\hat{\mathrm{k}}

\\\text { area of parallelogram }=|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=8 \sqrt{3} \text { . } \\ \overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & \alpha & 3 \\ 3 & -\alpha & 1 \end{array}\right|=\hat{\mathrm{i}}(4 \alpha)-\hat{\mathrm{j}}(-8)+\hat{\mathrm{k}}(-4 \alpha)

\\\therefore|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=\sqrt{64+32 \alpha^{2}}=8 \sqrt{3} \\ \Rightarrow 2+\alpha^{2}=6 \Rightarrow \alpha^{2}=4 \\ \therefore \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=3-\alpha^{2}+3=2

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