Evaluate a Third-Order Determinant
-The Area of a Triangle with Vertices

Multiple Choice

Evaluate a Third-Order Determinant
-The area of a triangle with vertices $\left( x _ { 1 } , y _ { 1 } \right) , \left( x _ { 2 } , y _ { 2 } \right)$ , and $\left( x _ { 3 } , y _ { 3 } \right)$ is
$$\text { Area } = \pm \frac { 1 } { 2 } \left| \begin{array} { l l l } x _ { 1 } & y _ { 1 } & 1 \\x _ { 2 } & y _ { 2 } & 1 \\x _ { 3 } & y _ { 3 } & 1\end{array} \right| \text {, }$$
where the symbol $\pm$ indicates that the appropriate sign should be chosen to yield a positive area. Use this formula to find the area of a triangle whose vertices are $( 6,10 ) , ( 8 , - 8 )$ , and $( - 9 , - 4 )$ .

A) 149
B) 298
C) 3
D) 6