The system of equations
$$-kx + 3y - 14z = 25$$
$$-15x + 4y - kz = 3$$
$$-4x + y + 3z = 4$$
is consistent for all $$k$$ in the set

The three linear equations can be written in matrix form as $$A\mathbf{x}= \mathbf{b}$$ where

$$A=\begin{bmatrix}-k & 3 & -14\\ -15 & 4 & -k\\ -4 & 1 & 3\end{bmatrix},\; \mathbf{x}=\begin{bmatrix}x\\y\\z\end{bmatrix},\; \mathbf{b}=\begin{bmatrix}25\\3\\4\end{bmatrix}.$$

A system of linear equations is
• always consistent when $$\det(A)\neq 0$$ (unique solution),
• possibly inconsistent when $$\det(A)=0$$ (then we must compare the ranks of $$A$$ and the augmented matrix $$[A\,|\,\mathbf{b}]$$).

First compute the determinant of $$A$$.

Using the first row for cofactor expansion,

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

Evaluate the three $$2\times 2$$ minors:

$$\begin{aligned} \begin{vmatrix}4 & -k\\1 & 3\end{vmatrix}&=4\cdot 3-(-k)\cdot 1=12+k,\\[4pt] \begin{vmatrix}-15 & -k\\-4 & 3\end{vmatrix}&=(-15)(3)-(-k)(-4)=-45-4k,\\[4pt] \begin{vmatrix}-15 & 4\\-4 & 1\end{vmatrix}&=(-15)(1)-4(-4)=-15+16=1. \end{aligned}$$

Substituting,

$$\det(A)=(-k)(12+k)-3(-45-4k)-14(1)$$ $$=-k^2-12k+135+12k-14$$ $$=121-k^2$$ $$=-(k^2-121).$$

Therefore

$$\det(A)=0 \Longleftrightarrow k^2-121=0 \Longleftrightarrow k=\pm 11.$$

Case 1: $$k\neq\pm 11$$.
Then $$\det(A)\neq 0$$, so $$A$$ is invertible and the system has a unique (hence consistent) solution.

Case 2: $$k=11$$.
Substitute $$k=11$$ in the first two equations:

$$\begin{aligned} -11x+3y-14z&=25\; -(1)\\ -15x+4y-11z&=3\; -(2) \end{aligned}$$

Multiply $$(1)$$ by $$4$$ and $$(2)$$ by $$3$$ to eliminate $$y$$:

$$\begin{aligned} -44x+12y-56z&=100,\\ -45x+12y-33z&=9. \end{aligned}$$

Subtract the second equation from the first:

$$x-23z=91\Longrightarrow x=91+23z.$$

Back in $$(1):\; -11(91+23z)+3y-14z=25$$ gives

$$y=342+89z.$$

Insert $$x,y$$ in the third original equation $$-4x+y+3z=4$$:

$$-4(91+23z)+(342+89z)+3z=4 \Longrightarrow -22=4,$$

a contradiction. Hence no solution; the system is inconsistent when $$k=11$$.

Case 3: $$k=-11$$.
The equations become

$$\begin{aligned} 11x+3y-14z&=25\; -(3)\\ -15x+4y+11z&=3\; -(4) \end{aligned}$$

Eliminate $$y$$ by multiplying $$(3)$$ by $$4$$ and $$(4)$$ by $$3$$:

$$\begin{aligned} 44x+12y-56z&=100,\\ -45x+12y+33z&=9. \end{aligned}$$

Subtract:

$$89x-89z=91\Longrightarrow x=z+\frac{91}{89}.$$

Substituting in $$(3)$$ gives $$y=z+\frac{408}{89}.$$

Use these in the third original equation $$-4x+y+3z=4$$:

$$-4\!\left(z+\frac{91}{89}\right)+\left(z+\frac{408}{89}\right)+3z =\frac{44}{89}\neq 4,$$

again a contradiction. Hence no solution; the system is inconsistent when $$k=-11$$.

Combining all cases, the system is consistent for every real $$k$$ except $$k=11$$ and $$k=-11$$.

Therefore the required set is $$\mathbb{R}-\{-11,11\}$$, which corresponds to Option D.