Step-by-step Solution:

Let the given determinant be defined as \( D = \begin{vmatrix} x & x+1 & x+3 \\ x+1 & x+3 & x+6 \\ x+3 & x+6 & x+10 \end{vmatrix} \). Apply row operations to simplify the matrix: \( R_2 \rightarrow R_2 - R_1 \) and \( R_3 \rightarrow R_3 - R_2 \) \[ D = \begin{vmatrix} x & x+1 & x+3 \\ 1 & 2 & 3 \\ 2 & 3 & 4 \end{vmatrix} \] Now, apply another row operation to introduce a zero: \( R_3 \rightarrow R_3 - 2R_2 \) \[ D = \begin{vmatrix} x & x+1 & x+3 \\ 1 & 2 & 3 \\ 0 & -1 & -2 \end{vmatrix} \] Expand the determinant along the simplified third row: \[ D = 0 - (-1) \cdot \left[ 3x - (x+3) \right] + (-2) \cdot \left[ 2x - (x+1) \right] \] \[ D = 1 \cdot (2x - 3) - 2 \cdot (x - 1) \] \[ D = 2x - 3 - 2x + 2 = -1 \] The determinant mathematically simplifies to a constant -1, which is entirely independent of the variable \( x \). Thus, even at \( x=2026 \), the value remains rigidly -1.