If the system of equations x + 5y + 6z = 4 , 2x + 3y + 4z = 7 , x +...

For the given system of equations to have infinitely many solutions, the determinant of the coefficient matrix must be zero, $\Delta = 0$.

$\Delta = \begin{vmatrix} 1 & 5 & 6 \\ 2 & 3 & 4 \\ 1 & 6 & a \end{vmatrix} = 0$

Expanding along the first row:

$1(3a - 24) - 5(2a - 4) + 6(12 - 3) = 0$

$3a - 24 - 10a + 20 + 54 = 0$

$-7a + 50 = 0 \Rightarrow a = \dfrac{50}{7}$

For infinitely many solutions, we must also have $\Delta_z = 0$.

$\Delta_z = \begin{vmatrix} 1 & 5 & 4 \\ 2 & 3 & 7 \\ 1 & 6 & b \end{vmatrix} = 0$

Expanding along the first row:

$1(3b - 42) - 5(2b - 7) + 4(12 - 3) = 0$

$3b - 42 - 10b + 35 + 36 = 0$

$-7b + 29 = 0 \Rightarrow b = \dfrac{29}{7}$

The point $(a, b)$ is \left(\dfrac{50}{7}, \dfrac{29}{7}\right)$.<br><br>Checking the given options, we evaluate$a - b$:<br><br>$a - b = \dfrac{50}{7} - \dfrac{29}{7} = \dfrac{21}{7} = 3$<br><br>Therefore, the point$(a, b)$lies on the line$x - y = 3$.<br><br>Answer:$x - y = 3