The area of the triangle whose vertices are (1, 1, 1) , (1, 2, 3), (2, 0, 1) is

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- $\sqrt{\frac{3}{2}}$

$\frac{1}{4}$

$\frac{3}{2}$

$\sqrt{\frac{3}{2}}$

answer is B.

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Detailed Solution

$\begin{array}{ll} \frac{1}{2} | \bar{AB} \times \bar{AC} | = \frac{1}{2} |\begin{matrix} \overline{i} & \overline{j} & \overline{k} \\ 0 & 1 & 2 \\ 1 & - 1 & 0 \end{matrix}| = \frac{1}{2} | i (2) - j (- 2) + k (- 1) | = \frac{1}{2} | 2 i + 2 j - \overline{k} | \\ = \frac{1}{2} \sqrt{4 + 4 + 1} = \frac{3}{2} \end{array}$

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