## Thomas' Calculus 13th Edition

$\sqrt{202}$
The first job is to find pairs of parallel vectors, so we are able to tell which two vectors that originate from a point (say, A) define the parallelogram. The three vectors with initial point A are $\overrightarrow{AB} = \langle 1-1, 7-0,2+1 \rangle= \langle 0, 7, 3 \rangle$ $\overrightarrow{AC} = \langle 1, 4, 0 \rangle$ $\overrightarrow{AD} = \langle-1, 3, 3 \rangle$ We search for parallel vectors $\overrightarrow{BC} = \langle 1, -3, -3 \rangle,$ which is parallel to $\overrightarrow{AD}$, $\overrightarrow{BD} = \langle-1, -4, 0 \rangle,$ which is parallel to $\overrightarrow{AC}$, We have our pairs of parallel vectors. The vectors defining the paralelogram are $\overrightarrow{AC}$ and $\overrightarrow{AD}$ ($\overrightarrow{AB}$ is the diagonal). Area =$| \overrightarrow{AC}\times \overrightarrow{AD} |$ $\overrightarrow{AC}\times\overrightarrow{AD}$=$\left|\begin{array}{lll} {\bf i} & {\bf j} & {\bf k}\\ 1 & 4 & 0\\ -1 & 3 & 3 \end{array}\right|=\langle (12-0) ,\ -(3-0) ,\ (3+4) \rangle$ $=\langle 12 ,\ -3 ,\ 7 \rangle$ Area =$| \overrightarrow{AB}\times\overrightarrow{AD} |$=$\sqrt{144+9+49}=\sqrt{202}$