To solve the determinant equation in problem 7:

Given matrix: $\begin{vmatrix} x+1 & x+2 & 3 \\ 3 & x+2 & x+1 \\ x+2 & 2 & x+3 \end{vmatrix} = 0$

We will expand this determinant along the first row and set it equal to zero to find the values of $x$.

The matrix expansion for the determinant is: $(x+1) \begin{vmatrix} x+2 & x+1 \\ 2 & x+3 \end{vmatrix} - (x+2) \begin{vmatrix} 3 & x+1 \\ x+2 & x+3 \end{vmatrix} + 3 \begin{vmatrix} 3 & x+2 \\ x+2 & 2 \end{vmatrix} = 0$

Now, we evaluate each of these smaller 2x2 determinants individually and substitute back.

1. For the first term: $(x+2)(x+3) - (x+1)(2) = x^2 + 5x + 6 - 2x - 2 = x^2 + 3x + 4$

2. For the second term: $3(x+3) - (x+1)(x+2) = 3x + 9 - (x^2 + 3x + 2) = -x^2 + 6$