Answer:
Let \[\overrightarrow{OA},\] \[\overrightarrow{OB}\] and \[\overrightarrow{OC}\] be the position vectors of points A, B and C respectively. Then, \[\overrightarrow{OA}=2\hat{i}+3\hat{j}+5\hat{k},\] \[\overrightarrow{OB}=3\hat{i}+5\hat{j}+8\hat{k}\] and \[\overrightarrow{OC}=2\hat{i}+7\hat{j}+8\hat{k}\] Then, \[\overrightarrow{AB}=\] Position vector of \[B-\] Position vector of A \[=\overrightarrow{OB}-\overrightarrow{OA}=3\hat{i}+5\hat{j}+8\hat{k}-(2\hat{i}+3\hat{j}+5\hat{k})\] \[=\hat{i}+2\hat{j}+3\hat{k}\] and \[\overrightarrow{AC}=\] Position vector of \[C-\] Position vector of A \[=\overrightarrow{OC}-\overrightarrow{OA}=2\hat{i}+7\hat{j}+8\hat{k}-(2\hat{i}+3\hat{j}+5\hat{k})\] \[=4\hat{j}+3\hat{k}\] Now, area of \[\Delta ABC=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|\,\,=\frac{1}{2}\left| \left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 1 & 2 & 3 \\ 0 & 4 & 3 \\ \end{matrix} \right| \right|\] \[=\frac{1}{2}|i(6-12)-\hat{j}(3-0)+\hat{k}(4-0)|\] \[=\frac{1}{2}|-\,6\hat{i}-3\hat{j}+4\hat{k}|\] \[=\frac{1}{2}\sqrt{{{(-\,6)}^{2}}+{{(-\,3)}^{2}}+{{(4)}^{2}}}\] \[=\frac{1}{2}\sqrt{36+9+16}=\frac{1}{2}\sqrt{61}\] Hence, area of \[\Delta ABC\] is \[\frac{1}{2}\sqrt{61}\] sq. units
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