 Solution

The second-largest eigenvalue is $2$, with eigenvector $(1,-1,0{)}^{\mathsf{T}}$. A unit principal component loading is thereforeIt is orthogonal to the first loading $u_1=(1,1,\sqrt{2}{)}^{\mathsf{T}}/2$. Either sign of a loading is equivalent, provided its principal component scores use the same sign convention.