First read the question and then we'll answer gradually.
Given for some scalars, not all , means $\exists c_i :c_i \not=0$ for $1 \leq i \leq n$
$\sum_{i=1}^{n}c_i.a_i=0$ where $a_i$ are column vectors (of size nx1) in $R_n$
so this means
$c_1\begin{bmatrix} x_{1,1}\\ x_{2,1}\\ .\\ .\\ .\\ x_{n,1} \end{bmatrix}$ +$c_2\begin{bmatrix} x_{1,2}\\ x_{2,2}\\ .\\ .\\ .\\ x_{n,2} \end{bmatrix}+....+$$c_n\begin{bmatrix} x_{1,n}\\ x_{2,n}\\ .\\ .\\ .\\ x_{n,n} \end{bmatrix}=0$
so if I make a vector of all $c_i$ for $1 \leq i \leq n$, such that say vector $k=$$\begin{bmatrix} c_1\\ c_2\\ .\\ .\\ .\\ c_n \end{bmatrix}$, this $k$ would be a non-zero vector.
Now, given $A=[a_1...a_n]$, so I have a vector k such that $Ak=0$
or in other way, $|A-0.I|K=0$ and 0 is an eigenvalue of this matrix and hence, this matrix A has determinant 0.
Okay, now since $Ak=0$, for any constant $h \not=0,A(hk)=0$ and I choose the domain of h as over all real numbers except 0.
Now, given that $b=\sum_{i=1}^{n}a_i$ which means you sum up all the columns of A and you would get b
so
$A.\begin{bmatrix} 1\\ 1\\ 1\\ .\\ .\\ 1 \end{bmatrix}=b$ and let this nx1 dimensional vector of all 1's be say $x$, so $Ax=b$
Now, $Ax=b$ and $A(hk)=0$ imply $A(x+hk)=b$ for any-non zero constant h.
Hence, this system of equations has infinite solutions because your h is infinite.
Answer-(C)