Look at \(L\) geometrically rather than through its matrix. The map \(L(x,y,z)=(x,y,0)\) drops the \(z\)-coordinate to zero and keeps \(x\) and \(y\) unchanged, so it is the orthogonal projection of \(\mathbb{R}^3\) onto the \(xy\)-plane.
For statement (I): the image of a projection onto the \(xy\)-plane is the \(xy\)-plane itself, a 2-dimensional subspace of \(\mathbb{R}^3\). So \(\operatorname{rank}(L)=2\), confirming (I).
For statements (II) and (III), use the defining property of a projection: \(L\) fixes every vector already lying in the \(xy\)-plane, and it sends every vector along the \(z\)-axis to the zero vector.
Any nonzero vector \((x,y,0)\) in the \(xy\)-plane satisfies \(L(x,y,0) = (x,y,0)\), so it is an eigenvector with eigenvalue 1. This plane is 2-dimensional, giving a full 2-dimensional family of independent eigenvectors with eigenvalue 1, for instance \((1,0,0)\) and \((0,1,0)\).
Any nonzero vector \((0,0,z)\) along the \(z\)-axis satisfies \(L(0,0,z) = (0,0,0) = 0\cdot(0,0,z)\), so it is an eigenvector with eigenvalue 0. This gives one more independent eigenvector, say \((0,0,1)\).
No other eigenvalues are possible for a projection, because projections satisfy \(L^2 = L\), which forces every eigenvalue \(\lambda\) to satisfy \(\lambda^2=\lambda\), that is \(\lambda \in \{0,1\}\). This confirms (II).
Combining the two eigenvalue-1 directions with the one eigenvalue-0 direction gives 3 linearly independent eigenvectors in total, namely \((1,0,0),(0,1,0),(0,0,1)\), not 2. So statement (III), which claims only 2 independent eigenvectors, is false.
\[\boxed{\text{Only (I) and (II) are true}}\]