Step 1: Recall that vectors $v_1,\dots,v_k$ form a linearly independent set iff the only solution to $a_1v_1+\cdots+a_kv_k=0$ is $a_1=\cdots=a_k=0$. Apply this to $c_1=Me_1$ and $c_2=Me_2$, the first two columns of $M$.
Step 2: For (I), rank$(M)=1$ means every column lies in a 1-dimensional subspace, i.e., all columns are scalar multiples of some fixed vector $v$ (or are zero). So $c_1=\alpha v$ and $c_2=\beta v$ for scalars $\alpha,\beta$. Any two vectors lying in a 1-dimensional space are automatically linearly dependent, regardless of $\alpha,\beta$, so $\{c_1,c_2\}$ is dependent. Hence (I) is false.
Step 3: For (II), exhibit a concrete rank-2 example directly through its columns: let $c_1=c_2=(0,1,0,0)^T$ and $c_3=(1,0,0,1)^T$ be the three columns of $M$, so $Me_1=c_1$, $Me_2=c_2$, $Me_3=c_3$. Since $c_1$ and $c_3$ are linearly independent and $c_2=c_1$, the column space of $M$ is spanned by $c_1$ and $c_3$, a 2-dimensional space, so rank$(M)=2$. But $c_1=c_2$ makes $\{Me_1,Me_2\}$ linearly dependent, so (II) can fail; it is not true in general.
Step 4: For (III), rank$(M)=3$ gives $\dim(\ker M) = 3-3=0$ by the rank-nullity theorem, so $M$ is injective on $\mathbb{R}^3$. An injective linear map carries linearly independent sets to linearly independent sets, and $\{e_1,e_2\}$ is independent in $\mathbb{R}^3$, so $\{Me_1,Me_2\}$ is independent in $\mathbb{R}^4$. So (III) is always true.
Step 5: Only (III) holds in every case. \[\boxed{\text{Only (III)}}\]