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Explanation:

Area (∆) = $\left|\begin{array}{ccc}{x}_1& y_1& 1\\ x_2& y_2& 2\\ x_3& y_3& 3\end{array}\right|$

where x₁ = y₁ = a, x₂ = a + 1, y₂ = a, x₃ = a and y₃ = a + 2

∴ Area (∆) = $\frac{1}{2}\left|\begin{array}{ccc}a& a& 1\\ a+1& a& 1\\ a& a+12& 1\end{array}\right|$

∴ Area (∆) = $\frac{1}{2}${a[a(1) - (a + 2)(1)] - a[(a + 1)(1) - a(1)] + 1[(a + 1)(a + 2) - a²]} = $\frac{1}{2}$ {- 2a - a + 3a + 2} = 1

Hence, option (b).