If the following system of linear equations
$$2x + y + z = 5$$
$$x - y + z = 3$$
$$x + y + az = b$$
has no solution, then:

We are given the three simultaneous linear equations

$$\begin{aligned} 2x + y + z &= 5,\\ x - y + z &= 3,\\ x + y + a\,z &= b. \end{aligned}$$

The nature of the solution set of a system of three equations in three unknowns depends first of all on the determinant of the coefficient matrix. Let us write that coefficient matrix:

$$A=\begin{bmatrix} 2 & 1 & 1\\ 1 & -1 & 1\\ 1 & 1 & a \end{bmatrix}.$$

Its determinant is

$$\begin{aligned} \det(A) &= \begin{vmatrix} 2 & 1 & 1\\ 1 & -1 & 1\\ 1 & 1 & a \end{vmatrix}\\[4pt] &=2\left((-1)\,a-1\cdot1\right)\;-\;1\left(1\cdot a-1\cdot1\right)\;+\;1\left(1\cdot1-(-1)\cdot1\right)\\[4pt] &=2(-a-1)\;-\;(a-1)\;+\;(1+1)\\[4pt] &=-2a-2\;-\;a+1\;+\;2\\[4pt] &=-3a+1. \end{aligned}$$

For the system to have no solution, the coefficient matrix must be singular, i.e. its determinant must be zero, because a non-zero determinant would force a unique solution. Thus we put

$$-3a+1=0\quad\Longrightarrow\quad a=\frac13.$$

So inconsistency, if it occurs at all, can occur only when $$a=\dfrac13.$$ At that value of $$a$$ we examine whether the third equation is compatible with the first two.

Substituting $$a=\dfrac13$$ into the system we have

$$\begin{aligned} 2x + y + z &= 5, \quad -(1)\\ x - y + z &= 3, \quad -(2)\\ x + y + \dfrac13\,z &= b. \quad -(3) \end{aligned}$$

To see the relationship among the equations, we try to express Equation (3) as a linear combination of Equations (1) and (2). Let the combination be

$$\alpha(1)+\beta(2).$$

Matching the $$x$$-coefficients first gives

$$2\alpha+\beta=1. \quad -(i)$$

Matching the $$y$$-coefficients gives

$$\alpha-\beta=1. \quad -(ii)$$

Solving (i) and (ii) simultaneously, we add them to obtain

$$3\alpha=2\quad\Longrightarrow\quad\alpha=\frac23,$$

and substitute back to get

$$\beta=1-2\alpha=1-\frac43=-\frac13.$$

Now we check the $$z$$-coefficients:

$$\alpha(1)+\beta(1)=\frac23+\left(-\frac13\right)=\frac13,$$

which matches the $$\dfrac13$$ in Equation (3). Hence the left-hand side of Equation (3) is $$\dfrac23$$ times Equation (1) minus $$\dfrac13$$ times Equation (2). The corresponding right-hand side derived from that same combination is

$$\alpha(5)+\beta(3)=\frac23\cdot5-\frac13\cdot3=\frac{10}{3}-1=\frac73.$$

Therefore:

• If $$b=\dfrac73,$$ Equation (3) is exactly the same linear combination of the first two equations, so the entire system is consistent and has infinitely many solutions.
• If $$b\neq\dfrac73,$$ Equation (3) cannot be satisfied simultaneously with Equations (1) and (2). The system then becomes inconsistent and has no solution.

We have thus shown that the system lacks a solution precisely when

$$a=\frac13\quad\text{and}\quad b\neq\frac73.$$

Looking at the options, this condition corresponds to Option D.

Hence, the correct answer is Option D.