If $$y(x) = \begin{vmatrix} \sin x & \cos x & \sin x + \cos x + 1 \\ 27 & 28 & 27 \\ 1 & 1 & 1 \end{vmatrix}$$, $$x \in \mathbb{R}$$, then $$\frac{d^2y}{dx^2} + y$$ is equal to

The determinant is

$$y(x)= \begin{vmatrix} \sin x & \cos x & \sin x+\cos x+1\\ 27 & 28 & 27\\ 1 & 1 & 1 \end{vmatrix}$$

We simplify the determinant by a column operation. Replace the third column by $$C_3 \;-\; C_1 \;-\; C_2$$ (this does not change the value of a determinant):

$$ C_3 \longrightarrow C_3-C_1-C_2 \quad\Longrightarrow\quad y(x)= \begin{vmatrix} \sin x & \cos x & 1\\ 27 & 28 & -28\\ 1 & 1 & -1 \end{vmatrix} $$

Now expand along the first row:

$$ y(x)=\sin x\, \bigl(28\cdot(-1)-(-28)\cdot1\bigr) -\cos x\, \bigl(27\cdot(-1)-(-28)\cdot1\bigr) +1\, \bigl(27\cdot1-28\cdot1\bigr) $$

Simplifying each bracket:

$$ \begin{aligned} 28(-1)-(-28)(1) &= -28+28=0,\\ 27(-1)-(-28)(1) &= -27+28=1,\\ 27(1)-28(1) &= -1. \end{aligned} $$

Hence

$$y(x)=\sin x\cdot0-\cos x\cdot1+1\cdot(-1)=-\cos x-1.$$

Differentiate twice with respect to $$x$$.

First derivative: $$y'(x)=\frac{d}{dx}\bigl(-\cos x-1\bigr)=\sin x.$$

Second derivative: $$y''(x)=\frac{d}{dx}\bigl(\sin x\bigr)=\cos x.$$

Now compute $$y''+y$$:

$$y''(x)+y(x)=\cos x+\bigl(-\cos x-1\bigr)=-1.$$

Thus $$\displaystyle\frac{d^{2}y}{dx^{2}}+y=-1.$$

The expression is the constant $$-1$$ for all $$x\in\mathbb{R}$$. So the correct choice is Option A.