Let $$\lambda \in \mathbb{R}$$. The system of linear equations
$$2x_1 - 4x_2 + \lambda x_3 = 1$$
$$x_{1} - 6x_{2} + x_{3} = 2$$
$$\lambda x_1 - 10x_2 + 4x_3 = 3$$
is inconsistent for:

We have the three linear equations

$$$\begin{aligned} 2x_1-4x_2+\lambda x_3 &= 1,\\[2pt] x_1-6x_2+x_3 &= 2,\\[2pt] \lambda x_1-10x_2+4x_3 &= 3. \end{aligned}$$$

The coefficient matrix is

$$$A=\begin{bmatrix} 2 & -4 & \lambda\\ 1 & -6 & 1\\ \lambda & -10 & 4 \end{bmatrix},\qquad \text{and the augmented column is }\, \mathbf b=\begin{bmatrix}1\\2\\3\end{bmatrix}.$$$

A system of linear equations is inconsistent exactly when $$$\operatorname{rank}(A)\lt \operatorname{rank}\bigl([A\;|\;\mathbf b]\bigr).$$$ Whenever the determinant $$\det A\neq0$$ the rank of $$A$$ is 3, the augmented matrix also has rank 3, and the system is therefore consistent. Hence we first look for the values of $$\lambda$$ that make $$\det A=0.$$

Using the first row expansion formula $$$\det A =2\!\begin{vmatrix}-6&1\\-10&4\end{vmatrix} -(-4)\!\begin{vmatrix}1&1\\\lambda&4\end{vmatrix} +\lambda\!\begin{vmatrix}1&-6\\\lambda&-10\end{vmatrix},$$$ we compute each minor one by one:

$$$\begin{aligned} \begin{vmatrix}-6&1\\-10&4\end{vmatrix} &=(-6)(4)-1(-10)=-24+10=-14,\\[6pt] \begin{vmatrix}1&1\\\lambda&4\end{vmatrix} &=(1)(4)-(1)(\lambda)=4-\lambda,\\[6pt] \begin{vmatrix}1&-6\\\lambda&-10\end{vmatrix} &=(1)(-10)-(-6)(\lambda)=-10+6\lambda=6\lambda-10. \end{aligned}$$$

Substituting these values gives

$$$\det A =2(-14)+4(4-\lambda)+\lambda(6\lambda-10) =-28+16-4\lambda+6\lambda^2-10\lambda =6\lambda^2-14\lambda-12.$$$

Taking out a common factor 2,

$$\det A=2\bigl(3\lambda^2-7\lambda-6\bigr).$$

We now solve $$3\lambda^2-7\lambda-6=0.$$ The discriminant is

$$$\Delta=(-7)^2-4(3)(-6)=49+72=121=11^2,$$$ so

$$\lambda=\frac{7\pm11}{2\cdot3}=\frac{18}{6}\;\text{or}\;\frac{-4}{6},$$ that is,

$$\lambda=3\quad\text{or}\quad\lambda=-\dfrac23.$$

Thus $$\det A=0$$ only for these two values. For every other real $$\lambda$$ the system is consistent with a unique solution. We must now examine $$\lambda=3$$ and $$\lambda=-\dfrac23$$ separately to see whether the system becomes inconsistent.

Case $$\lambda=3$$: Substituting $$\lambda=3$$ gives

$$$\begin{aligned} 2x_1-4x_2+3x_3 &= 1,\\ x_1-6x_2+x_3 &= 2,\\ 3x_1-10x_2+4x_3 &= 3. \end{aligned}$$$

From the second equation we get $$x_1=2+6x_2-x_3.$$ Putting this in the first equation:

$$$2(2+6x_2-x_3)-4x_2+3x_3=1 \;\;\Longrightarrow\;\;4+12x_2-2x_3-4x_2+3x_3=1,$$$

which simplifies to

$$8x_2+x_3=-3\quad\Longrightarrow\quad x_3=-3-8x_2.$$ Inserting both expressions for $$x_1$$ and $$x_3$$ into the third equation:

$$3(2+6x_2-x_3)-10x_2+4x_3=3,$$

we get exactly $$8x_2+x_3=-3,$$ an identity that is already satisfied. Hence the third equation adds no new information, and there are infinitely many solutions. The system is consistent at $$\lambda=3$$.

Case $$\lambda=-\dfrac23$$: Substituting $$\lambda=-\dfrac23$$ gives

$$$\begin{aligned} 2x_1-4x_2-\dfrac23x_3 &= 1,\\ x_1-6x_2+x_3 &= 2,\\ -\dfrac23x_1-10x_2+4x_3 &= 3. \end{aligned}$$$

To clear fractions, multiply the first and third equations by 3:

$$$\begin{aligned} 6x_1-12x_2-2x_3 &= 3,\\ x_1-6x_2+x_3 &= 2,\\ -2x_1-30x_2+12x_3 &= 9. \end{aligned}$$$

From the middle equation we again have $$x_1=2+6x_2-x_3.$$ Substituting into the first:

$$$6(2+6x_2-x_3)-12x_2-2x_3=3 \;\;\Longrightarrow\;\; 12+36x_2-6x_3-12x_2-2x_3=3,$$$

which reduces to

$$$24x_2-8x_3=-9 \quad\Longrightarrow\quad -3x_2+x_3=\frac{9}{8}.$$$ Next, substitute $$x_1=2+6x_2-x_3$$ into the third equation:

$$-2(2+6x_2-x_3)-30x_2+12x_3=9,$$

giving

$$$-4-12x_2+2x_3-30x_2+12x_3=9 \;\;\Longrightarrow\;\; -42x_2+14x_3=13 \;\;\Longrightarrow\;\; -3x_2+x_3=\frac{13}{14}.$$$

We now see that the same combination $$-3x_2+x_3$$ is forced to equal two different numbers:

$$-3x_2+x_3=\frac{9}{8}\quad\text{and}\quad -3x_2+x_3=\frac{13}{14}.$$

Since $$\dfrac{9}{8}\neq\dfrac{13}{14},$$ no pair $$(x_2,x_3)$$ can satisfy both conditions simultaneously. Therefore the augmented matrix has rank 3 while the coefficient matrix has rank 2, and the system is inconsistent for $$\lambda=-\dfrac23.$$

We have found exactly one value of $$\lambda$$, namely the negative value $$-\dfrac23$$, that makes the system inconsistent. No positive value produces inconsistency, and there are not two such values.

Hence, the correct answer is Option B.