The even consecutive integers

$\begin{aligned} a^2 + b^2 + c^2 - ab-bc-ca & = a{\color{#3D99F6}(a-b)}+b{\color{#3D99F6}(b-c)}+c{\color{#D61F06}(c-a)} \\ & = {\color{#3D99F6}2}a+{\color{#3D99F6}2}b{\color{#D61F06}-4}c \\ & = 2a-2b+4b-4c \\ & = 2(a-b) + 4(b-c) \\ & = 2(2) + 4(2) \\ & = \boxed{12} \end{aligned}$

Here the one line solution $a^2+b^2+c^2-ab-bc-ca=\dfrac{1}{2} [(a-b)^2+(b-c)^2+(a-c)^2]=\dfrac{1}{2}[2^2+2^2+4^2]=\dfrac{1}{2}[4+4+16]=12$

Here is an alternate solution in case you do not understand any of the solutions here:

let the even consecutive integers be $2,4$ and $6$

then, $a=6,b=4$ and $c=2$

then we have,

$6^2+4^2+2^2-6(4)-4(2)-6(2)=36+16+4-24-8-12=$ $\boxed{12}$ $\large\color{#D61F06}\boxed{\text{answer}}$