If $$\left|\begin{array}{ccc}1 & 1 & 0 \\x^2+2 x+2 & 1 & 0 \\2 & 1 & 1\end{array}\right|=0$$​, then the value of $$x$$​ is

Correct option is D

​$$\begin{aligned}&\text{Let the matrix be:} \\&\Delta = \left|\begin{array}{ccc}1 & 1 & 0 \\x^2 + 2x + 2 & 1 & 1 \\2 & 1 & 1\end{array}\right| \\\\&\text{We can expand along the third column since it has two zeros, making the calculation easier.} \\\\&\Delta = 0 \cdot M_{13} - 1 \cdot M_{23} + 1 \cdot M_{33} \\\\&\text{Where } M_{ij} \text{ is the minor of element } (i, j), \text{ i.e., determinant of the } 2 \times 2 \text{ matrix formed by deleting the } i^{th} \text{ row} \\&\text{and } j^{th} \text{ column:} \\\\&M_{33} = \left|\begin{array}{cc}1 & 1 \\x^2 + 2x + 2 & 1\end{array}\right| = (1)(1) - (1)(x^2 + 2x + 2) = 1 - (x^2 + 2x + 2) \\&\qquad = 1 - x^2 - 2x - 2 = -x^2 - 2x - 1 \\\\&\text{Thus:} \\\\&\Delta = - (1 \cdot \left|\begin{array}{cc}1 & 1 \\2 & 1\end{array}\right|) + 1 \cdot \left|\begin{array}{cc}1 & 1 \\x^2 + 2x + 2 & 1\end{array}\right|) \\&\Delta = - (1 \cdot 1 - 1 \cdot 2) + (1 \cdot 1 - 1 \cdot (x^2 + 2x + 2)) \\&\Delta = - (1 - 2) + (1 - x^2 - 2x - 2) \\&\Delta = - (-1) + (-x^2 - 2x - 1) \\&\Delta = 1 - x^2 - 2x - 1 \\&\Delta = -x^2 - 2x \\\\&\text{Set determinant to zero:} \\\\&-x^2 - 2x = 0 \implies x^2 + 2x = 0 \implies x(x + 2) = 0 \implies x = 0 \text{ or } x = -2\end{aligned}$$​​