A) \[\sqrt{6}\]
B) \[2\sqrt{6}\]
C) \[3\sqrt{6}\]
D) \[4\sqrt{6}\]
Correct Answer: B
Solution :
\[\Delta =\sqrt{\Delta _{x}^{2}+\Delta _{y}^{2}+\Delta _{z}^{2}}\] where \[{{\Delta }_{x}}=\frac{1}{2}\left| \,\begin{matrix} {{y}_{1}} & {{z}_{1}} & 1 \\ {{y}_{2}} & {{z}_{2}} & 1 \\ {{y}_{3}} & {{z}_{3}} & 1 \\ \end{matrix}\, \right|\] and so on. Alter : Form two vectors \[\overrightarrow{AB}\] and \[\overrightarrow{AC}\] \[\Delta =\frac{1}{2}|\overrightarrow{AB}\,\times \,\overrightarrow{AC}|\] = \[\frac{1}{2}\left| \,\left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\ {{x}_{3}}-{{x}_{1}} & {{y}_{3}}-{{y}_{1}} & {{z}_{3}}-{{z}_{1}} \\ \end{matrix} \right|\, \right|\] = \[\frac{1}{2}\left| \,\left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & -3 \\ -1 & 3 & -1 \\ \end{matrix} \right|\, \right|\] = \[\frac{1}{2}|8\mathbf{i}+4\mathbf{j}+4\mathbf{k}|\] = \[\frac{1}{2}\sqrt{64+16+16}=\frac{\sqrt{96}}{2}=2\sqrt{6}\].
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