To solve for $y(x)$, we start with the determinant of the given matrix: $\begin{vmatrix} \sin x & \cos x & \sin x + \cos x + 1 \\ 27 & 28 & 27 \\ 1 & 1 & 1 \end{vmatrix}$ Using the formula for a $3 \times 3$ determinant, we expand along the first row: $f(x) = \sin x \cdot \begin{vmatrix} 28 & 27 \\ 1 & 1 \end{vmatrix} - \cos x \cdot \begin{vmatrix} 27 & 27 \\ 1 & 1 \end{vmatrix} + (\sin x + \cos x + 1) \cdot \begin{vmatrix} 27 & 28 \\ 1 & 1 \end{vmatrix}$ Calculating the smaller $2 \times 2$ determinants: $\begin{vmatrix} 28 & 27 \\ 1 & 1 \end{vmatrix} = 28 \cdot 1 - 27 \cdot 1 = 1$ $\begin{vmatrix} 27 & 27 \\ 1 & 1 \end{vmatrix} = 27 \cdot 1 - 27 \cdot 1 = 0$ $\begin{vmatrix} 27 & 28 \\ 1 & 1 \end{vmatrix} = 27 \cdot 1 - 28 \cdot 1 = -1$ Substituting these into the determinant calculation gives: $f(x) = \sin x \cdot 1 - \cos x \cdot 0 + (\sin x + \cos x + 1) \cdot (-1)$ $f(x) = \sin x - (\sin x + \cos x + 1)$ $f(x) = \sin x - \sin x - \cos x - 1$ $f(x) = -\cos x - 1$ Now, calculate the derivatives: First derivative $f'(x)$: $f'(x) = \frac{d}{dx}(-\cos x - 1) = \sin x$ Second derivative $\frac{d^2 f}{dx^2}$: $\frac{d^2 f}{dx^2} = \frac{d}{dx}(\sin x) = \cos x$ Finally, compute $\frac{d^2 y}{dx^2} + y$: $\frac{d^2 f}{dx^2} + f(x) = \cos x - \cos x - 1 = -1$ Thus, $\frac{d^2 y}{dx^2} + y$ is equal to $-1$.