solution

Using row operations we can reduce the matrix $A$ to the following matrix which has the same determinat as $A$ ( Why?).

$\vert det \left( \left[ \begin{array}{lll} a_1& a_2 & 1 \\ b_1 &b_2 & 1\\ c_... ...}\right] \right)\vert = \vert( b_1- a_1)(c_2 -a_2)-(b_2 -a_2)(c_1 -a_1) \vert$

Since $AB = ( b_1 -a_1, b_2 - a_2)$ and $AC= (c_1- a_1 , c_2-a_2)$, the area of the parallelogram formed by $AB$ and $AC$ is

$\vert det \left( \left[ \begin{array}{ll} b_1- a_1 &b_2 -a_2 \\ c_1 -a_1&c_... ...}\right] \right)\vert = \vert( b_1- a_1)(c_2 -a_2)-(b_2 -a_2)(c_1 -a_1) \vert$