The area of the parallelogram generated by $\overrightarrow{OA}$ and $\overrightarrow{OB}$ is given by
\[\|\overrightarrow{OA} \times \overrightarrow{OB}\| = \left\| \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \times \begin{pmatrix} -3 \\ -2 \\ 1 \end{pmatrix} \right\| = \left\| \begin{pmatrix} 8 \\ -10 \\ 4 \end{pmatrix} \right\| = 6 \sqrt{5}.\][asy]
unitsize(0.4 cm);

pair A, B, C, D;

A = (0,0);
B = (7,2);
C = (1,3);
D = B + C;

draw(A--B,Arrow(6));
draw(A--C,Arrow(6));
draw(B--C);
draw(B--D--C,dashed);

label("$O$", A, SW);
label("$A$", B, SE);
label("$B$", C, W);
[/asy]

Therefore, the area of triangle $OAB$ is $\boxed{3 \sqrt{5}}.$
Question:
The points $O,$ $A,$ and $B$ satisfy $\overrightarrow{OA} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ and $\overrightarrow{OB} = \begin{pmatrix} -3 \\ -2 \\ 1 \end{pmatrix}.$  Find the area of triangle $OAB.$
Answer:
Let's think step by step.

user
3 \sqrt{5}
The points $O,$ $A,$ and $B$ satisfy $\overrightarrow{OA} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}$ and $\overrightarrow{OB} = \begin{pmatrix} -3 \\ -2 \\ 1 \end{pmatrix}.$  Find the area of triangle $OAB.$
test