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\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
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\begin{document}


{\bf Question}

\begin{description}
\item[(a)] Find the area of the triangle at $(1,2,3), \, (3,2,1), \,
(2,3,1)$.
\item[(b)] Find the equation of the plane containing the above
triangle.
\end{description}

\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)] $A = (1,2,3), \, B = (3,2,1), \, C = (2,3,1)$.

$\vec{AB} = (2, 0, -2) \, \vec{AC}=(1,1-2) \, \vec{AB}
\times\vec{AC} = (2, 2, 2)$

Area of triangle $ABC = \frac{1}{2} |AB \times AC| =
\frac{1}{2}\sqrt{12} = \sqrt 3$
\item[(b)] $AB \times AC$ is normal to the plane $ABC,$ so its
equation is $2x + 2y + 2z = k$.  It contains $(1, 2, 3)$ so $k =
12$

Therefore $x+y+z = 6$
\end{description}


\end{document}