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(C) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is given by the formula:Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(

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$a = 9$ or $a = -1$

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(C) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2),$ and $(x_3, y_3)$ is given by the formula:Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$Given vertices are $A(a, 5), B(6, 7),$ and $C(2, 3)$ with Area $= 10$.Substituting the values into the formula:$10 = \frac{1}{2} |a(7 - 3) + 6(3 - 5) + 2(5 - 7)|$$20 = |a(4) + 6(-2) + 2(-2)|$$20 = |4a - 12 - 4|$$20 = |4a - 16|$This implies two cases:Case $1$: $4a - 16 = 20 \implies 4a = 36 \implies a = 9$Case $2$: $4a - 16 = -20 \implies 4a = -4 \implies a = -1$Thus,the possible values for $a$ are $9$ or $-1$.

If the area of the triangle having the vertices $A (a, 5), B (6, 7), C (2, 3)$ is $10$,then find $a$.

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