If $$S$$ is the set of distinct values of $$b$$ for which the following system of linear equations
$$x + y + z = 1$$
$$x + ay + z = 1$$
$$ax + by + z = 0$$
has no solution, then $$S$$ is:

We have the three simultaneous linear equations

$$\begin{aligned} x+y+z &= 1 \qquad\qquad (1)\\ x+ay+z &= 1 \qquad\qquad (2)\\ ax+by+z &= 0 \qquad\qquad (3) \end{aligned}$$

Our task is to locate all real numbers $$b$$ for which this system has no solution.

To see when inconsistency arises, we first compare the first two equations. Subtracting equation $$ (2)$$ from equation $$ (1)$$ gives

$$ (x-x)\;+\;(y-ay)\;+\;(z-z)\;=\;1-1, $$

so

$$ (1-a)\,y = 0. \quad -(4) $$

This single relation already produces two mutually exclusive situations. We study each of them separately.

Case I : $$a\neq1$$

Because $$1-a\neq0$$, equation $$ (4)$$ forces

$$ y=0. \quad -(5) $$

Substituting $$y=0$$ into equation $$ (1)$$ we obtain

$$ x+0+z = 1 \;\;\Longrightarrow\;\; x+z = 1. \quad -(6) $$

Next we substitute $$y=0$$ into equation $$ (3)$$. The term $$by$$ vanishes, leaving

$$ ax+z = 0. \quad -(7) $$

Now we have the two linear equations in the two unknowns $$x$$ and $$z$$:

$$ \begin{cases} x+z = 1,\\[2pt] ax+z = 0. \end{cases} $$

Subtracting the second from the first eliminates $$z$$ and yields

$$ (1-a)\,x = 1. \quad -(8) $$

Because $$a\neq1$$, the coefficient $$1-a$$ is non-zero, so equation $$ (8)$$ gives the unique value

$$ x = \dfrac{1}{1-a}. \quad -(9) $$

Putting this value back into equation $$ (6)$$, we get

$$ \frac{1}{1-a}+z = 1 \;\;\Longrightarrow\;\; z = 1-\frac{1}{1-a} = \frac{-a}{1-a} = \frac{a}{a-1}. \quad -(10) $$

Thus for every real number $$b$$ (no restriction at all), when $$a\neq1$$ the system produces the explicit solution

$$ (x,y,z) = \left(\frac{1}{1-a},\;0,\;\frac{a}{a-1}\right), $$

so the system is perfectly consistent. Hence, whenever $$a\neq1$$ no value of $$b$$ can destroy solvability.

Case II : $$a=1$$

Setting $$a=1$$ in the original system, equations $$ (1)$$ and $$ (2)$$ become identical:

$$ x+y+z = 1. \quad -(11) $$

Equation $$ (3)$$, after inserting $$a=1$$, turns into

$$ x+by+z = 0. \quad -(12) $$

Thus the system effectively reduces to the pair of equations $$ (11)$$ and $$ (12)$$. To compare these, subtract equation $$ (12)$$ from equation $$ (11)$$:

$$ (x-x)\;+\;(y-by)\;+\;(z-z)\;=\;1-0, $$

which simplifies to

$$ (1-b)\,y = 1. \quad -(13) $$

Now again two possibilities appear:

• If $$b\neq1$$, equation $$ (13)$$ delivers the single value

$$ y = \dfrac{1}{1-b}, \quad -(14) $$

and we can solve for $$x+z$$ from either equation $$ (11)$$ or $$ (12)$$. Indeed using $$ (11)$$ we get $$x+z=1-y$$, while using $$ (12)$$ we get $$x+z=-by$$, and with $$y$$ from $$ (14)$$ the two expressions coincide, so solutions exist. Therefore no inconsistency occurs when $$b\neq1$$.

• If $$b=1$$, equation $$ (13)$$ reads $$0\cdot y = 1$$, an impossibility. Thus the two equations $$ (11)$$ and $$ (12)$$ demand the same left-hand side but force it to equal two different right-hand sides (1 and 0). This is a direct contradiction, so the system has no solution when $$a=1$$ and $$b=1$$.

Collecting the conclusions

• For every $$a\neq1$$ the system is solvable for all real $$b$$.

• For $$a=1$$ the system is solvable for every $$b\neq1$$ but becomes inconsistent precisely at $$b=1$$.

Hence the lone real number capable of destroying solvability is

$$ b = 1. $$

Therefore the set $$S$$ of such $$b$$ is the singleton $$\{1\}$$, containing exactly one element.

Hence, the correct answer is Option D.