= Solution
In a real orthonormal <basis>, a <matrix> is a proper <rotation matrix> exactly when $R^TR=I$ and $\det R=1$. The first condition preserves <dot products>, and the second preserves <orientation of a vector space>.
Put $s=\sqrt3$, $p=(1+s)/3$, $q=(1-s)/3$ and $r=1/3$, so the rows of the given <matrix> are $(p,q,r),(r,p,q),(q,r,p)$. Their squared <Euclidean norms> and pairwise <dot products> are
$$
p^2+q^2+r^2=1,\qquad pq+qr+rp=0.
$$
Thus $R^TR=I$. For the sign of its <determinant>, use the circulant form:
$$
\det R=p^3+q^3+r^3-3pqr
=(p+q+r)(p^2+q^2+r^2-pq-qr-rp)=1.
$$
This proves that it is a <rotation matrix>. Its fixed axis is $(1,1,1)^T$, because each row sums to one. The <rotation trace formula> gives $\cos\theta=(\operatorname{tr}R-1)/2=\sqrt3/2$; the skew part selects a positive <angle> $\theta=\pi/6$ about the unit axis $(1,1,1)^T/\sqrt3$.
The original triangle has edge <vectors> $b-a=(-2,2,0)^T$ and $c-a=(-2,0,2)^T$. Its <area> is half the <Euclidean norm> of their <cross product>:
$$
\boxed{\mathcal A=\tfrac12\|(b-a)\times(c-a)\|=\tfrac12\|(4,4,4)^T\|=2\sqrt3.}
$$
The rotated vertices are twice the three columns of $R$:
$$
\boxed{a'=\frac23\begin{pmatrix}1+\sqrt3\\1\\1-\sqrt3\end{pmatrix},\qquad
b'=\frac23\begin{pmatrix}1-\sqrt3\\1+\sqrt3\\1\end{pmatrix},\qquad
c'=\frac23\begin{pmatrix}1\\1-\sqrt3\\1+\sqrt3\end{pmatrix}.}
$$
A <rotation matrix> preserves lengths and the <angle> between the two edge <vectors>, so it preserves their parallelogram <area>. More explicitly, $(Ru)\times(Rv)=R(u\times v)$ for a proper <rotation matrix>, and $R(1,1,1)^T=(1,1,1)^T$, so the rotated edge <cross product> is again $(4,4,4)^T$. Therefore
$$
\boxed{\mathcal A'=2\sqrt3.}
$$
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