• # question_answer The area of the triangle whose vertices are at the points (2, 1, 1), (3, 1, 2) and (-4, 0, 1) is A)  $\sqrt{19}$                B)  $\frac{1}{2}\sqrt{19}$ C)  $\frac{1}{2}\sqrt{38}$                     D)  $\frac{1}{2}\sqrt{57}$

$\because$    ${{\Delta }_{x}}=\frac{1}{2}\,\left| \begin{matrix} {{y}_{1}} & {{z}_{1}} & 1 \\ {{y}_{2}} & {{z}_{2}} & 1 \\ {{y}_{3}} & {{z}_{3}} & 1 \\ \end{matrix} \right|$ $=\frac{1}{2}\,\left| \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 1 \\ 0 & 1 & 1 \\ \end{matrix} \right|=\frac{1}{2}$ ${{\Delta }_{y}}=\frac{1}{2}\,\left| \begin{matrix} {{z}_{1}} & {{x}_{1}} & 1 \\ {{z}_{2}} & {{x}_{2}} & 1 \\ {{z}_{3}} & {{x}_{3}} & 1 \\ \end{matrix} \right|$ $=\frac{1}{2}\,\left| \begin{matrix} 1 & 2 & 1 \\ 2 & 3 & 1 \\ 1 & -4 & 1 \\ \end{matrix} \right|=-3$ ${{\Delta }_{z}}=\frac{1}{2}\,\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\ {{x}_{2}} & {{y}_{2}} & 1 \\ {{x}_{3}} & {{y}_{3}} & 1 \\ \end{matrix} \right|$ $=\frac{1}{2}\,\left| \begin{matrix} 2 & 1 & 1 \\ 3 & 1 & 1 \\ -4 & 0 & 1 \\ \end{matrix} \right|=-\frac{1}{2}$ $\therefore$    Area of triangle $=\sqrt{\Delta _{x}^{2}+\Delta _{y}^{2}+\Delta _{z}^{2}}$ $=\sqrt{\frac{1}{4}+9+\frac{1}{4}}=\frac{1}{2}\sqrt{38}$