 Solution

In a real orthonormal basis, a matrix is a proper rotation matrix exactly when $R^{T}R=I$ and $\det R=1$. The first condition preserves dot products, and the second preserves orientation of a vector space.

Put $s=\sqrt{3}$, $p=(1+s)/3$, $q=(1-s)/3$ 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 areThus $R^{T}R=I$. For the sign of its determinant, use the circulant form: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 $\cos \theta =(\mathrm{tr}R-1)/2=\sqrt{3}/2$; the skew part selects a positive angle $\theta =\pi /6$ about the unit axis $(1,1,1{)}^T/\sqrt{3}$.

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:The rotated vertices are twice the three columns of $R$: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