If the system of equations

x + 2y + z = 5

2x + y + αz = 5

8x + y + z = 18 has no solution, then α is equal to

The correct option is 9/15.

To find the value of $\alpha$ for which the given system of linear equations has no solution, we can use Cramer's Rule (determinant method).

The given system of equations is:
1) $x+2y+z=5$
2) $2x+y+\alpha z=5$
3) $8x+y+z=18$

For a system of three linear equations in three variables to have no solution, the determinant of the coefficient matrix, denoted as $D$, must be equal to zero, and at least one of the determinants $D_x$, $D_y$, or $D_z$ must be non-zero.

Let's first write down the determinant of the coefficient matrix $D$:
$D=\left|\begin{array}{ccc}1& 2& 1\\ 2& 1& \alpha \\ 8& 1& 1\end{array}\right|$

We expand this determinant along the first row:
$D=1·(1·1-\alpha ·1)-2·(2·1-\alpha ·8)+1·(2·1-1·8)$

Simplifying the terms inside the parentheses:
$D=1(1-\alpha )-2(2-8\alpha )+1(2-8)$

Expanding the expression:
$D=1-\alpha -4+16\alpha -6$

Combining the like terms:
$D=15\alpha -9$

For the system to have no solution, we set $D=0$:
$15\alpha -9=0$
$15\alpha =9$
$\alpha =\frac{9}{15}$

To confirm that this value indeed results in no solution (and not infinitely many solutions), we check the determinant $D_x$ by replacing the first column of $D$ with the constant terms on the right-hand side of the equations:
$D_x=\left|\begin{array}{ccc}5& 2& 1\\ 5& 1& \frac{9}{15}\\ 18& 1& 1\end{array}\right|$

Substituting $\frac{9}{15}=\frac{3}{5}$ and expanding along the first row:
$D_x=5·(1-\frac{3}{5})-2·(5-18·\frac{3}{5})+1·(5-18)$
$D_x=5·\left(\frac{2}{5}\right)-2·(5-\frac{54}{5})-13$
$D_x=2-2·(-\frac{29}{5})-13$
$D_x=2+\frac{58}{5}-13=-11+11.6=0.6\ne 0$

Since $D=0$ and $D_x\ne 0$ when $\alpha =\frac{9}{15}$, the system is inconsistent and has no solution.