If the system of equations
$$x - 2y + 3z = 9$$
$$2x + y + z = b$$
$$x - 7y + az = 24$$
has infinitely many solutions, then $$a - b$$ is equal to __________

For a system of three linear equations in the three unknowns $$x,\;y,\;z$$ to possess infinitely many solutions, the three equations must be linearly dependent. In other words, one of the equations must be obtainable as a linear combination of the other two, and this same combination must reproduce the corresponding constant term on the right hand side. We therefore assume that suitable real numbers $$c_1$$ and $$c_2$$ exist such that

$$ c_1\,(x-2y+3z) \;+\; c_2\,(2x+y+z) \;=\; x-7y+az $$

and simultaneously

$$ c_1\,(9)\;+\;c_2\,(b)\;=\;24 . $$

We now equate coefficients of $$x,\;y,\;z$$ one by one. Starting with the $$x$$-coefficients, we have

$$ c_1\cdot1\;+\;c_2\cdot2 \;=\;1 , \qquad\text{so } c_1+2c_2=1. \quad -(1) $$

Next, comparing the $$y$$-coefficients gives

$$ c_1\cdot(-2)\;+\;c_2\cdot1 \;=\;-7, \qquad\text{so } -2c_1+c_2=-7. \quad -(2) $$

From equations (1) and (2) we solve for $$c_1$$ and $$c_2$$. Using (1) to express $$c_1$$, we get

$$ c_1 = 1-2c_2. $$

Substituting this in (2) yields

$$ -2(1-2c_2)+c_2 = -7 \;\Longrightarrow\; -2 + 4c_2 + c_2 = -7 \;\Longrightarrow\; 5c_2 = -5 \;\Longrightarrow\; c_2 = -1. $$

Putting $$c_2=-1$$ back into $$c_1=1-2c_2$$ gives

$$ c_1 = 1 - 2(-1) = 1+2 = 3. $$

Now we equate the $$z$$-coefficients. From the assumed dependence we have

$$ c_1\cdot3 + c_2\cdot1 = a, \qquad\text{i.e. } 3c_1 + c_2 = a. \quad -(3) $$

Substituting $$c_1=3$$ and $$c_2=-1$$ into (3) gives

$$ 3(3) + (-1) = 9 - 1 = 8, $$

so

$$ a = 8. $$

Finally, the constant terms must also match, so we impose

$$ c_1\cdot9 + c_2\cdot b = 24. \quad -(4) $$

Substituting $$c_1=3$$ and $$c_2=-1$$ in (4) yields

$$ 3\cdot9 + (-1)\cdot b = 24 \;\Longrightarrow\; 27 - b = 24 \;\Longrightarrow\; b = 3. $$

Therefore

$$ a-b = 8-3 = 5. $$

So, the answer is $$5$$.