Now how do I do this?

Relevant wiki: Linear Transformations

Area of triangles, $\Delta = z = \begin{vmatrix} b&c&1\\c&a&1\\a&b&1 \end{vmatrix}$

$\Delta_1 = \begin{vmatrix} ac-b^2& ab-c^2&1\\ba-c^2& bc-a^2&1\\cb-a^2&ca-b^2&1 \end{vmatrix}$

Now let's define a matrix A, $\begin{bmatrix} b&c&1\\c&a&1\\a&b&1 \end{bmatrix}\cdot A = \begin{bmatrix} ac-b^2& ab-c^2&1\\ba-c^2& bc-a^2&1\\cb-a^2&ca-b^2&1 \end{bmatrix}$

Applying Row Transformation, $R_1 \to R_1-R_3\\R_2 \to R_2-R_3$

$\begin{bmatrix} b-a&c-b&0\\c-a&a-b&0\\a&b&1 \end{bmatrix}\cdot A = (a+b+c)\begin{bmatrix} a-b& c-b&0\\c-a& b-a&0\\\dfrac{cb-a^2}{a+b+c}&\dfrac{ca-b^2}{a+b+c}&\dfrac1{a+b+c} \end{bmatrix}$

Taking determinant, $\mathrm{det}(A) = (a+b+c)^2\\\Rightarrow \dfrac{\Delta_1}{\Delta}=(a+b+c)^2$

For $z$ , $z = 3abc-a^3-b^3-c^3 = -(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

So, the final answer turns out to be $\mathrm{Ans.} = \dfrac12 \left[(a-b)^2+(b-c)^2+(c-a)^2\right]$

Numerically, $\color{#3D99F6}{\dfrac{4+9+25}2 = \boxed{19}}$