Consider a matrix $A_{3X4}$ = $\begin{bmatrix} 1 & 3& 3 &2 \\ 2& 6 & 9&7 \\ -1 &-3 &3 & 4 \end{bmatrix}$
The Echelon form of this matrix after Row operations
$R_2=R_2-2R_1$
$R_3=R_3+R_1$
$R_3=R_3-2R_2$
would be
$\begin{bmatrix} 1 & 3& 3 &2 \\ 0& 0 & 3&3\\ 0 &0 &0 & 0 \end{bmatrix}$
Now, we have our system $Ax=b$
which can be represented as
$\begin{bmatrix} 1 & 3& 3 &2 \\ 2& 6 & 9&7 \\ -1 &-3 &3 & 4 \end{bmatrix}$.$\begin{bmatrix} u\\ v\\ w\\ y \end{bmatrix}=\begin{bmatrix} b_1\\ b_2\\ b_3 \end{bmatrix}$
Perform the same Row operations as listed above and then it becomes
$\begin{bmatrix} 1 & 3 & 3 & 2\\ 0&0 & 3 & 3\\ 0&0 & 0 & 0 \end{bmatrix}.\begin{bmatrix} u\\ v\\ w\\ y \end{bmatrix}=\begin{bmatrix} b_1\\ b_2-2b_1\\ b_3-2b_2+5b_1 \end{bmatrix}$
Let this system be $Ux=c$
Now, your system would be consistent, if $b_3-2b_2+5b_1=0$ which validates option (A).
Option (B), says if $Ax=0$ then this system has infinitely many solutions and yes, because you take any constant $c \not=0$,
$Acx=0$
Option(D) says if A is an nxn matrix and rank(A)=n, means all the rows and columns of A are linearly independent, then there is no such non-zero vector that can take some combinations of columns of A and produce a zero vector. And, thus the only solution to $Ax=0$ is the vector $X=0$ which is a trivial solution, so this option is true.
Option(C) asserts that if m=n and b is a non-zero vector, then system has unique solution which is clearly false.
Why?
For the system, $Ax=b$ to have a solution for a non-zero b, the vector b must be such that it can be produced by some linear combinations of columns of A.This means, b must be in the Column Space of A.
In the above system $Ux=c$ it has a solution when $c=$$\begin{bmatrix} 1(b_1))\\ 5(b_2)\\ 5(b_3) \end{bmatrix}$ and it is inconsistent, means does not have a solution when $c=$$\begin{bmatrix} 1(b_1))\\ 5(b_2)\\ 6(b_3) \end{bmatrix}$
Hence, Option (C) is the answer.