If the system of equations x + y + z = 5, x + 2y + 3z = 9, x + 3y +...

The system of linear equations ${a}_{1}x + {b}_{1}y + {c}_{1}z + {d}_{1} = 0, {a}_{2}x + {b}_{2}y + {c}_{2}z + {d}_{2} = 0$ and ${a}_{3}x + {b}_{3}y + {c}_{3}z + {d}_{3} = 0$ have infinitely many solutions, if

D = \left|\begin{cases}{a}_{1} & {b}_{1} & {c}_{1} \\ {a}_{2} & {b}_{2} & {c}_{2} \\ {a}_{3} & {b}_{3} & {c}_{3}\end{cases}\right| = 0, {D}_{1} = \left|\begin{cases}{d}_{1} & {b}_{1} & {c}_{1} \\ {d}_{2} & {b}_{2} & {c}_{2} \\ {d}_{3} & {b}_{3} & {c}_{3}\end{cases}\right| = 0, {D}_{2} = \left|\begin{cases}{a}_{1} & {d}_{1} & {c}_{1} \\ {a}_{2} & {d}_{2} & {c}_{2} \\ {a}_{3} & {d}_{3} & {c}_{3}\end{cases}\right| = 0

and {D}_{3} = \left|\begin{cases}{a}_{1} & {b}_{1} & {d}_{1} \\ {a}_{2} & {b}_{2} & {d}_{2} \\ {a}_{3} & {b}_{3} & {d}_{3}\end{cases}\right| = 0 .

Hence, for the system of equations $x + y + z = 5, x + 2y + 3z = 9$ and $x + 3y + \alphaz = \beta$ have infinitely many solution, if

D = \left|\begin{cases}1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & \alpha\end{cases}\right| = 0

$⇒ \left(2\alpha - 9\right) - \left(\alpha - 3\right) + \left(3 - 2\right)$

$⇒ 2\alpha - 9 - \alpha + 3 + 1 = 0$

$⇒ \alpha = 5 .$

And, {D}_{1} = \left|\begin{cases}5 & 1 & 1 \\ 9 & 2 & 3 \\ \beta & 3 & \alpha\end{cases}\right| = 0

⇒ \left|\begin{cases}5 & 1 & 1 \\ 9 & 2 & 3 \\ \beta & 3 & 5\end{cases}\right| = 0

$⇒ 5\left(10 - 9\right) - \left(45 - 3\beta\right) + \left(27 - 2\beta\right) = 0$

$⇒ 5 - 45 + 3\beta + 27 - 2\beta = 0$

$⇒ \beta = 13 .$

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