It is extremely easy. Your answer is absolutely correct.
We have to reduce the rows into the row echelon form according to the Gauss-Jordan Reduction or elimination method to obtain the zeroes in the following first two columns and obtaining the solution with the assistance of an augmented matrix. Here is my process to get the D) Option or the fourth option. Using the interpreter LaTeX:
[tex]\begin{vmatrix}3 & -4 & 2 \\ \\ 0 & -1 & 7 \\ \\ \dfrac{3}{2} & -2 & 1 \end{vmatrix}[/tex]
Perform sets of operations to augment the matrix equation of the given sets by row operations with scalar products.
[tex]\mathbf{R_3 \longleftarrow R_3 - \dfrac{1}{2} \times R_1}[/tex]
[tex]\begin{vmatrix}3 & -4 & 2 \\ 0 & -1 & 7 \\ 0 & 0 & 0 \end{vmatrix}[/tex]
[tex]\mathbf{R_2 \longleftarrow - 1 \times R_2}[/tex]
[tex]\begin{vmatrix}3 & -4 & 2 \\ 0 & 1 & -7 \\ 0 & 0 & 0 \end{vmatrix}[/tex]
[tex]\mathbf{R_1 \longleftarrow R_1 + 4 \times R_2}[/tex]
[tex]\begin{vmatrix}3 & 0 & -26 \\ 0 & 1 & -7 \\ 1 & 0 & 0 \end{vmatrix}[/tex]
[tex]\mathbf{R_1 \longleftarrow \dfrac{1}{3} \times R_1}[/tex]
[tex]\boxed{\underline{\begin{vmatrix}1 & 0 & -\dfrac{26}{3} \\ \\ 0 & 1 & -7 \\ \\ 0 & 0 & 0 \end{vmatrix}}}[/tex]
Hope it helps.
