Let c1 ….. cn be scalars, not all zero, such that \[\sum\limits_{{\text{i}} = 1}^{\text{n}} {{{\text{c}}_{\text{i}}}{{\text{a}}_{\text{i}}}} = 0\] where ai are column vectors in Rn. Consider the set of linear equations Ax = b where A = [a1 ….. an] and \[{\text{b}} = \sum\limits_{{\text{i}} = 1}^{\text{n}} {{{\text{a}}_{\text{i}}}} .\] The set of equations has A. a unique solution at x = Jn where Jn denotes a n-dimensional vector of all 1 B. no solution C. infinitely many solutions D. finitely many solutions

The correct answer is $\boxed{\text{C}}$.

The set of linear equations $Ax = b$ has infinitely many solutions if and only if the augmented matrix $[A|b]$ has more rows than columns. In this case, the augmented matrix is $$\left[\begin{array}{cccc} a_1 & a_2 & \cdots & a_n \\ c_1 a_1 & c_2 a_2 & \cdots & c_n a_n \\ \vdots & \vdots & \ddots & \vdots \\ c_1 a_n & c_2 a_n & \cdots & c_n a_n \end{array}\right].$$ Since $c_1, c_2, \dots, c_n$ are not all zero, the determinant of this matrix is nonzero. Therefore, the augmented matrix has full rank and the system of equations has infinitely many solutions.

To see that the solution is not unique, note that any vector $x$ of the form $$x = \sum_{i=1}^n c_i a_i$$ is a solution to the system of equations. Therefore, there are infinitely many solutions.

The answer $\boxed{\text{A}}$ is incorrect because the set of equations has infinitely many solutions, not a unique solution.

The answer $\boxed{\text{B}}$ is incorrect because the set of equations has infinitely many solutions, not no solution.

The answer $\boxed{\text{D}}$ is incorrect because the set of equations has infinitely many solutions, not finitely many solutions.