A) \[7\sqrt{10}\,sq\]units
B) \[\frac{1}{2}\sqrt{10}\,\text{sq}\,\text{units}\]
C) \[\frac{7}{2}\sqrt{10}\,\text{sq}\,\text{units}\]
D) None of these
Correct Answer: C
Solution :
Let \[{{A}_{x}}{{A}_{y}}\]and \[{{A}_{z}}\]be the areas of projection of \[\Delta \Alpha \Beta C\] on \[yz,zx\]and \[xy-\]planes. Then, \[{{A}_{x}}=\frac{1}{2}\times \text{Absolute}\,\text{values}\,\text{of}\,\left| \begin{matrix} 2 & 3 & 1 \\ -1 & 1 & 1 \\ 2 & -4 & 1 \\ \end{matrix} \right|=\frac{21}{2}\] \[{{A}_{y}}=\frac{1}{2}\times \text{Absolute}\,\text{values}\,\text{of}\,\left| \begin{matrix} 3 & 1 & 1 \\ 1 & 2 & 1 \\ -4 & 1 & 1 \\ \end{matrix} \right|=\frac{7}{2}\] and \[{{A}_{z}}=\frac{1}{2}\times \text{Absolute}\,\text{values}\,\text{of}\,\left| \begin{matrix} 1 & 2 & 1 \\ 2 & -1 & 1 \\ 1 & 2 & 1 \\ \end{matrix} \right|\] \[=0\] \[\therefore \]Area of \[\Delta \Alpha \Beta C=\sqrt{{{A}_{x}}^{2}+{{A}_{y}}^{2}+{{A}_{z}}^{2}}\] \[=\sqrt{\frac{441}{4}+\frac{49}{4}+0}=\frac{7}{2}\sqrt{10}\,\text{sq}\,\text{units}\]
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