We interpret the chained statement as the system

$$-kx+3y-14z=25,$$ $$-15x+4y-kz=3,$$ $$-4x+y+3z=4.$$

We must find all values of $k$ for which this system is consistent.

1. Write the coefficient matrix

The coefficient matrix is

$$A=\begin{pmatrix} -k & 3 & -14\\ -15 & 4 & -k\\ -4 & 1 & 3 \end{pmatrix}$$

and the constant column is

$$B=\begin{pmatrix}25\\3\\4\end{pmatrix}.$$

A linear system is certainly consistent whenever $\det(A)\neq 0$. So first compute $\det(A)$.

2. Compute the determinant

Expanding along the first row:

$$\det(A)=(-k)\begin{vmatrix}4 & -k\\1 & 3\end{vmatrix}-3\begin{vmatrix}-15 & -k\\-4 & 3\end{vmatrix}+(-14)\begin{vmatrix}-15 & 4\\-4 & 1\end{vmatrix}.$$

Now evaluate each minor:

$$\begin{vmatrix}4 & -k\\1 & 3\end{vmatrix}=12+k,$$ $$\begin{vmatrix}-15 & -k\\-4 & 3\end{vmatrix}=-45-4k,$$ $$\begin{vmatrix}-15 & 4\\-4 & 1\end{vmatrix}=-15+16=1.$$

Hence

$$\det(A)=(-k)(12+k)-3(-45-4k)-14.$$

Simplifying,

$$\det(A)=-12k-k^2+135+12k-14=121-k^2.$$

So

$$\det(A)=121-k^2=(11-k)(11+k).$$

Thus $\det(A)=0$ when

$$k=11 \quad \text{or} \quad k=-11.$$

For all other real $k$, the system has a unique solution, hence is consistent.

3. Check the exceptional values

We must check whether the system is still consistent at $k=\pm 11$.

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Case 1: $k=11$

The system becomes

$$-11x+3y-14z=25 \quad ...(1)$$ $$-15x+4y-11z=3 \quad ...(2)$$ $$-4x+y+3z=4 \quad ...(3)$$

From $(3)$,

$$y=4+4x-3z.$$

Substitute into $(1)$:

$$-11x+3(4+4x-3z)-14z=25$$ $$-11x+12+12x-9z-14z=25$$ $$x-23z=13. \quad ...(4)$$

Substitute into $(2)$:

$$-15x+4(4+4x-3z)-11z=3$$ $$-15x+16+16x-12z-11z=3$$ $$x-23z=-13. \quad ...(5)$$

Equations $(4)$ and $(5)$ contradict each other. Hence the system is inconsistent for $k=11$.

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Case 2: $k=-11$

The system becomes

$$11x+3y-14z=25 \quad ...(1)$$ $$-15x+4y+11z=3 \quad ...(2)$$ $$-4x+y+3z=4 \quad ...(3)$$

From $(3)$,

$$y=4+4x-3z.$$

Substitute into $(1)$:

$$11x+3(4+4x-3z)-14z=25$$ $$11x+12+12x-9z-14z=25$$ $$23x-23z=13$$ $$x-z=\frac{13}{23}. \quad ...(4)$$

Substitute into $(2)$:

$$-15x+4(4+4x-3z)+11z=3$$ $$-15x+16+16x-12z+11z=3$$ $$x-z=-13. \quad ...(5)$$

Again $(4)$ and $(5)$ contradict. Hence the system is inconsistent for $k=-11$.

4. Final set of values

Therefore, the system is consistent for all real $k$ except

$$k=11, -11.$$

So the required set is

$$\mathbb{R}-\{-11,11\}.$$

5. Match with options

This corresponds to Option D.