Consistency of Linear System and Determinant Condition

To determine the consistency of the system of linear equations, we write the system in matrix form:

$$\begin{array}{rcccl} x &-& 2y &+& 3z = -1 \\ -x &+& y &-& 2z = k \\ x &-& 3y &+& 4z = 1 \end{array}$$

First, let us evaluate the determinant of the coefficient matrix, $D$:

$$D = \begin{vmatrix} 1 & -2 & 3 \\ -1 & 1 & -2 \\ 1 & -3 & 4 \end{vmatrix}$$

Applying the elementary row operations $R_2 \to R_2 + R_1$ and $R_3 \to R_3 - R_1$:

$$D = \begin{vmatrix} 1 & -2 & 3 \\ 0 & -1 & 1 \\ 0 & -1 & 1 \end{vmatrix} = 1((-1)(1) - (1)(-1)) = 0$$

Since $D = 0$, by Cramer's Rule, the system of equations will have:

1. Infinitely many solutions if $D_x = D_y = D_z = 0$.
2. No solution (inconsistent) if at least one of $D_x, D_y, D_z \neq 0$.

Now, let us evaluate the determinant given in STATEMENT-2:

$$\Delta = \begin{vmatrix} 1 & 3 & -1 \\ -1 & -2 & k \\ 1 & 4 & 1 \end{vmatrix}$$

Expanding along the first row:
$\Delta = 1((-2)(1) - (k)(4)) - 3((-1)(1) - (k)(1)) - 1((-1)(4) - (-2)(1))$
$\Delta = 1(-2 - 4k) - 3(-1 - k) - 1(-4 + 2)$
$\Delta = -2 - 4k + 3 + 3k + 2 = 3 - k$

Thus, $\Delta \neq 0$ for $k \neq 3$.
Hence, STATEMENT-2 is True.

Notice that the determinant $\Delta$ is related to $D_y$ (the determinant obtained by replacing the column of coefficients of $y$ with the column of constants):

$$D_y = \begin{vmatrix} 1 & -1 & 3 \\ -1 & k & -2 \\ 1 & 1 & 4 \end{vmatrix}$$

Interchanging Column 2 and Column 3 yields:

$$D_y = - \begin{vmatrix} 1 & 3 & -1 \\ -1 & -2 & k \\ 1 & 4 & 1 \end{vmatrix} = -\Delta = k - 3$$

Since $D = 0$ and $D_y = k - 3 \neq 0$ for $k \neq 3$, at least one of the determinants $D_x, D_y, D_z$ is non-zero.
Therefore, the given system of linear equations is inconsistent and has no solution for $k \neq 3$.

Hence, STATEMENT-1 is True, and STATEMENT-2 is the correct explanation for STATEMENT-1.

Therefore, the correct option is (A).