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Solution

Let

$D=\left|\begin{array}{ccc}x & x+1 & x+3\\ x+1 & x+3 & x+6\\ x+3 & x+6 & x+10\end{array}\right|$

Apply row operations:

$R_2\to R_2-R_1$

$R_3\to R_3-R_1$

Then,

$D=\left|\begin{array}{ccc}x & x+1 & x+3\\ 1 & 2 & 3\\ 3 & 5 & 7\end{array}\right|$

Now expanding along the first row,

$D=x(2\cdot 7-3\cdot 5)-(x+1)(1\cdot 7-3\cdot 3)+(x+3)(1\cdot 5-2\cdot 3)$

$D=x(14-15)-(x+1)(7-9)+(x+3)(5-6)$

$D=-x+2(x+1)-(x+3)$

$D=-x+2x+2-x-3$

$D=-1$

So, at $x=2026$, the value of the determinant is $-1$.