Proving the area is irrational for triangle with integer vertices

I assume that the plane on which the points lay is normal to the vector $(1,1,1)$. After a translation, we can assume that one of the points is $(0,0,0)$ and the other two have integer coordinates: let them be $(a,b,c)$ and $(d,e,f)$. Notice that the area of the triangle can be seen as the determinant of the matrix having as columns $(a,b,c)$, $(d,e,f)$ and $(1,1,1)$, divided by twice the length of $(1,1,1)$, since it is normal to the other two points (the idea is that you are calculating the volume of a solid having as base a quadrilateral having twice the area you are interested in). The determinant of a matrix with integer coefficients is integer, and the length of $(1,1,1)$ is $\sqrt{3}$. So the area is irrational.

Let $\vec{AB}(a,b,c)$ and $\vec{AC}(p,q,r).$

Thus, $\{a,b,c,p,q,r\}\subset\mathbb Z$, $a+b+c=p+q+r=0$ and $$S_{\Delta ABC}=\frac{1}{2}\sqrt{a^2+b^2+c^2}\cdot\sqrt{p^2+q^2+r^2}\cdot\sqrt{1-\left(\frac{ap+bq+cr}{\sqrt{a^2+b^2+c^2}\cdot\sqrt{p^2+q^2+r^2}}\right)^2}=$$ $$=\frac{1}{2}\sqrt{(a^2+b^2+c^2)(p^2+q^2+r^2)-(ap+bq+cr)^2}=$$ $$=\frac{1}{2}\sqrt{4(a^2+ab+b^2)(p^2+pq+q^2)-(ap+bq+(a+b)(p+q))^2}=$$ $$=\frac{1}{2}\sqrt{3(aq-bp)^2}=\frac{\sqrt3}{2}|aq-bp|\not\in\mathbb Q$$ and we are done!

We'll show that if three non-collinear points on the given plane have rational coordinates, then the area of the triangle with those points as vertices is irrational.

Suppose $A',B',C'$ are three non-collinear points with rational coordinates on the given plane.

Let $G$ be the centroid of triangle $A'B'C'$.

Then $G = {\large{\frac{A'+B'+C'}{3}}}$, hence $G$ also has rational coordinates.

Let $A = A'-G,\;\;B=B'-G,\;\;C=C'-G$.

Then $A,B,C$ are nonzero (as vectors), have rational coordinates, and reside in the plane with equation $$x+y+z = 0$$ so the normal vector of the plane is still the vector $n=\langle{1,1,1}\rangle$.

Since triangle $ABC$ is just a translation of triangle $A'B'C'$, the area stays the same.

Thus, it suffices to show the area of triangle $ABC$ is irrational.

But now we have $A + B + C = 0$, so $A = -B-C$.

Let $k\;$denote the area of triangle $ABC$. \begin{align*} \text{Then}\;\;k &={\small{\frac{1}{2}}}|AB\times AC|\\[4pt] &={\small{\frac{1}{2}}}|(B-A)\times (C-A)|\\[4pt] &={\small{\frac{1}{2}}}|(2B+C)\times (2C+B)|\qquad\text{[since$\;A = -B - C$]}\\[4pt] &={\small{\frac{1}{2}}}|4B{\times}C\,+\,2B{\times}B\,+\,2C{\times}C\,+\,C{\times}B|\\[4pt] &={\small{\frac{1}{2}}}|4B{\times}C\,+\,C{\times}B|\\[4pt] &={\small{\frac{1}{2}}}|3B{\times}C|\\[4pt] &={\small{\frac{3}{2}}}|B{\times}C|\\[4pt] \end{align*} so it suffices to show $|B\times C|$ is irrational.

But $B\times C$ is a nonzero vector with rational coordinates, and is parallel to $n$.

It follows that $B\times C = \langle{t,t,t}\rangle$, for some nonzero rational number $t$, hence $|B\times C| = \sqrt{3t^2} = |t|\sqrt{3},\;$which is irrational.