For which of the following ordered pairs $$(\mu, \delta)$$, the system of linear equations
$$x + 2y + 3z = 1$$
$$3x + 4y + 5z = \mu$$
$$4x + 4y + 4z = \delta$$
is inconsistent?

We have the system of three linear equations

$$\begin{aligned} x+2y+3z&=1 \\[2pt] 3x+4y+5z&=\mu \\[2pt] 4x+4y+4z&=\delta \end{aligned}$$

To test whether the system is consistent or inconsistent we compare the rank of the coefficient matrix with the rank of the augmented matrix. If the augmented matrix has a higher rank than the coefficient matrix, the system is inconsistent.

First, let us write the coefficient matrix $$A$$ and find its determinant to know its rank.

$$A=\begin{bmatrix} 1&2&3\\ 3&4&5\\ 4&4&4 \end{bmatrix}$$

The determinant of a $$3\times3$$ matrix $$\begin{vmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33} \end{vmatrix}$$ is given by $$a_{11}\begin{vmatrix}a_{22}&a_{23}\\a_{32}&a_{33}\end{vmatrix} -a_{12}\begin{vmatrix}a_{21}&a_{23}\\a_{31}&a_{33}\end{vmatrix} +a_{13}\begin{vmatrix}a_{21}&a_{22}\\a_{31}&a_{32}\end{vmatrix}.$$

Applying this formula,

$$\begin{aligned} \det(A)&= 1\begin{vmatrix}4&5\\4&4\end{vmatrix} -2\begin{vmatrix}3&5\\4&4\end{vmatrix} +3\begin{vmatrix}3&4\\4&4\end{vmatrix}\\[6pt] &=1(4\cdot4-5\cdot4)-2(3\cdot4-5\cdot4)+3(3\cdot4-4\cdot4)\\[6pt] &=1(16-20)-2(12-20)+3(12-16)\\[6pt] &=(-4)-2(-8)+3(-4)\\[6pt] &=-4+16-12\\[6pt] &=0. \end{aligned}$$

Because $$\det(A)=0$$, the rows of $$A$$ are linearly dependent and the rank of the coefficient matrix is at most $$2$$.

Let us find the exact linear relation among the three rows to see how the constants must relate for consistency. Denote the three rows of $$A$$ by

$$R_1=(1,2,3),\qquad R_2=(3,4,5),\qquad R_3=(4,4,4).$$

Assume that $$R_3=aR_1+bR_2$$. We solve for $$a$$ and $$b$$ using the first two components.

From the first component: $$a\cdot1+b\cdot3=4\quad\Longrightarrow\quad a+3b=4.$$

From the second component: $$a\cdot2+b\cdot4=4\quad\Longrightarrow\quad 2a+4b=4.$$

Dividing the second equation by $$2$$ gives $$a+2b=2$$. Now subtract:

$$\bigl(a+3b\bigr)-\bigl(a+2b\bigr)=4-2\quad\Longrightarrow\quad b=2.$$

Substituting $$b=2$$ into $$a+2b=2$$ yields $$a+4=2$$, so $$a=-2$$.

Therefore,

$$R_3=-2R_1+2R_2.$$

For consistency, the same relation must hold for the constants in the right-hand column. The constants are $$1$$ (from the first equation), $$\mu$$ (from the second), and $$\delta$$ (from the third). Hence we require

$$-2\cdot1+2\cdot\mu=\delta.$$

Simplifying, we obtain the consistency condition

$$\boxed{\;\delta=2\mu-2\;}.$$

If $$\delta\neq2\mu-2$$, then the augmented matrix attains rank $$3$$ whereas the coefficient matrix has rank $$2$$; consequently the system is inconsistent.

Now we check each option:

$$\textbf{(A)}\;(\mu,\delta)=(4,3):\; 2\mu-2=2\cdot4-2=6\neq3\;\Rightarrow$$ inconsistent.

$$\textbf{(B)}\;(\mu,\delta)=(4,6):\; 2\mu-2=6=6\;\Rightarrow$$ consistent.

$$\textbf{(C)}\;(\mu,\delta)=(1,0):\; 2\mu-2=0=0\;\Rightarrow$$ consistent.

$$\textbf{(D)}\;(\mu,\delta)=(3,4):\; 2\mu-2=4=4\;\Rightarrow$$ consistent.

Only the ordered pair in Option A violates the consistency condition, so the system is inconsistent precisely for that choice.

Hence, the correct answer is Option A.