4
A ball is thrown upwards from the plane surface of the ground.
Suppose the plane surface from which the ball is thrown consists of the points
$A(1,0,2),\; B(3,-1,1)$ and $C(1,2,1)$ on it.
The highest point of the ball takes is $D(2,3,1)$ as shown in the figure.
Using this information answer the question.

The area of $\triangle ABC$ is:
Solution
$A(1,0,2)$
$B(3,-1,1)$
$C(1,2,1)$
Vectors
$\vec{AB}=(3-1,-1-0,1-2)=(2,-1,-1)$
$\vec{AC}=(1-1,2-0,1-2)=(0,2,-1)$
Cross product
$\vec{AB}\times\vec{AC}$
$=
\begin{vmatrix}
\mathbf{i}&\mathbf{j}&\mathbf{k}\\
2&-1&-1\\
0&2&-1
\end{vmatrix}
$
$=3\mathbf{i}+2\mathbf{j}+4\mathbf{k}$
Magnitude
$|\vec{AB}\times\vec{AC}|$
$=\sqrt{3^2+2^2+4^2}$
$=\sqrt{29}$
Area of triangle
$=\frac{1}{2}|\vec{AB}\times\vec{AC}|$
$=\frac{1}{2}\sqrt{29}$