Which one is correct, following system of linear equations 2x – 3y + 4z = 7, 3x – 4y + 5z = 8, 4x – 5y + 6z = 9 has?

Solving the given system of equation by Cramer?s rule, we get,
x = D₁/D, y = D₂/D, z = D₃/D where,
D = \(\begin{vmatrix}2 & -3 & 4 \\3 & -4 & 5 \\4 & -5 & 6 \end {vmatrix}\)
D = –\(\begin{vmatrix}2 & 3 & 4 \\3 & 4 & 5 \\4 & 5 & 6 \end {vmatrix}\)
Now, performing, C₃ = C₃ ? C₂ and C₂ = C₂ ? C₁ we get,
D = –\(\begin{vmatrix}2 & 1 & 1 \\3 & 1 & 1 \\4 & 1 & 1 \end {vmatrix}\)
As two columns have identical values, so,
D = 0
Similarly,
D₁ = \(\begin{vmatrix}7 & -3 & 4 \\8 & -4 & 5 \\9 & -5 & 6 \end {vmatrix}\)
Now, performing, C₁ = C₁ ? C₃
D₁ = –\(\begin{vmatrix}3 & -3 & 4 \\3 & -4 & 5 \\3 & -5 & 6 \end {vmatrix}\)
Now, performing, C₃ = C₃ ? C₂
D₁ = -3\(\begin{vmatrix}1 & -3 & 1 \\1 & -4 & 1 \\1 & -5 & 1 \end {vmatrix}\)
As two columns have identical values, so,
D₁ = 0
D₂ = \(\begin{vmatrix}2 & 7 & 4 \\3 & 9 & 5 \\4 & 8 & 6 \end {vmatrix}\)
Now, performing,
D₂ = –\(\begin{vmatrix}2 & 3 & 2 \\3 & 3 & 2 \\4 & 3 & 2 \end {vmatrix}\)
Now, performing, C₂ = C₂ ? C₃ and C₃ = C₃ ? C₁
D₂ = 6\(\begin{vmatrix}2 & 1 & 1 \\3 & 1 & 1 \\4 & 1 & 1 \end {vmatrix}\)
As two columns have identical values, so,
D₂ = 0
D₃ = \(\begin{vmatrix}2 & -3 & 7 \\3 & -4 & 9 \\4 & -5 & 6 \end {vmatrix}\)
D₃ = –\(\begin{vmatrix}2 & 3 & 4 \\3 & 4 & 4 \\4 & 5 & 4 \end {vmatrix}\)
Now, performing, C₂ = C₂ ? C₂ and C₃ = C₃ ? C₂
D₃ = -4\(\begin{vmatrix}2 & 1 & 1 \\3 & 1 & 1 \\4 & 1 & 1 \end {vmatrix}\)
As two columns have identical values, so,
D₃ = 0
Since, D = D₁ = D₂ = D₃ = 0, thus, it has infinitely many solutions.