If the system of equations $$x + y + z = 5$$, $$x + 2y + 3z = 9$$, $$x + 3y + \alpha z = \beta$$ has infinitely many solutions, then $$\beta - \alpha$$ equals:

We have the three simultaneous linear equations

$$\begin{aligned} x + y + z &= 5,\\ x + 2y + 3z &= 9,\\ x + 3y + \alpha z &= \beta. \end{aligned}$$

For a system of three equations in three unknowns to possess infinitely many solutions, the rank of the coefficient matrix must be less than the number of variables while still equalling the rank of the augmented matrix. A necessary first step is therefore that the determinant of the coefficient matrix must vanish.

The coefficient matrix is

$$A=\begin{bmatrix} 1 & 1 & 1\\ 1 & 2 & 3\\ 1 & 3 & \alpha \end{bmatrix}.$$

We evaluate its determinant. Using the formula for the determinant of a $$3\times3$$ matrix

$$\det A = a_{11}(a_{22}a_{33}-a_{23}a_{32}) - a_{12}(a_{21}a_{33}-a_{23}a_{31}) + a_{13}(a_{21}a_{32}-a_{22}a_{31}),$$

and substituting $$a_{ij}$$ from $$A$$ we get

$$\det A = 1\bigl(2\alpha - 3\!\cdot\!3\bigr) - 1\bigl(1\alpha - 3\!\cdot\!1\bigr) + 1\bigl(1\!\cdot\!3 - 2\!\cdot\!1\bigr).$$

Simplifying each term one by one:

$$\begin{aligned} 1(2\alpha - 9) &= 2\alpha - 9,\\ -1(\alpha - 3) &= -\alpha + 3,\\ 1(3 - 2) &= 1. \end{aligned}$$

Adding these three expressions, we obtain

$$\det A = (2\alpha - 9) + (-\alpha + 3) + 1 = \alpha - 5.$$

For infinitely many solutions we set $$\det A = 0,$$ so

$$\alpha - 5 = 0 \;\Longrightarrow\; \alpha = 5.$$

Now that the determinant is zero, we must also ensure that the augmented matrix has the same rank as the coefficient matrix. This means the third equation must be a linear combination of the first two. Let us therefore assume that there exist scalars $$\lambda$$ and $$\mu$$ such that

$$\lambda\,(x + y + z = 5) + \mu\,(x + 2y + 3z = 9) = x + 3y + 5z = \beta.$$

Equating coefficients of like terms gives four equations:

From the $$x$$-coefficients: $$\lambda + \mu = 1.$$

From the $$y$$-coefficients: $$\lambda + 2\mu = 3.$$

From the $$z$$-coefficients: $$\lambda + 3\mu = 5.$$

From the constants: $$5\lambda + 9\mu = \beta.$$

We first solve for $$\lambda$$ and $$\mu$$ using the simplest pair. Subtracting the first relation from the second:

$$(\lambda + 2\mu) - (\lambda + \mu) = 3 - 1 \;\Longrightarrow\; \mu = 2.$$

Substituting $$\mu = 2$$ into $$\lambda + \mu = 1$$ we find

$$\lambda + 2 = 1 \;\Longrightarrow\; \lambda = -1.$$

We verify that these values also satisfy the $$z$$-coefficient condition: $$\lambda + 3\mu = -1 + 3\!\cdot\!2 = -1 + 6 = 5,$$ which matches the required coefficient of $$z$$, so the choice is consistent.

Now we compute $$\beta$$ from the constants’ relation:

$$\beta = 5\lambda + 9\mu = 5(-1) + 9(2) = -5 + 18 = 13.$$

Finally, we need the value of $$\beta - \alpha.$$ Substituting $$\beta = 13$$ and $$\alpha = 5$$ we get

$$\beta - \alpha = 13 - 5 = 8.$$

Hence, the correct answer is Option A.