To find the condition for the system of equations to have infinitely many solutions, we can use the method of row reduction on the augmented matrix.

Augmented Matrix Row Reduction

The given system of equations is:

$x + 5y + 6z = 4$

$2x + 3y + 4z = 7$

$x + 6y + az = b$

The augmented matrix is:

Perform row operations to simplify:

1. $R_2 \rightarrow R_2 - 2R_1$
2. $R_3 \rightarrow R_3 - R_1$

The matrix becomes:

Swap $R_2$ and $R_3$ for easier calculation:

Perform the next row operation:

3. $R_3 \rightarrow R_3 + 7R_2$

The matrix becomes:

Simplify the last row:

Coefficient of $z$: $-8 + 7a - 42 = 7a - 50$

Constant term: $-1 + 7b - 28 = 7b - 29$

The final row-reduced matrix is:

Condition for Infinite Solutions

For the system to have infinitely many solutions, the last row must represent the equation $0z = 0$. This requires both the coefficient of $z$ and the constant term in the last row to be zero.

$7a - 50 = 0 \implies a = \frac{50}{7}$

$7b - 29 = 0 \implies b = \frac{29}{7}$

Therefore, the point $(a, b)$ is $(\frac{50}{7}, \frac{29}{7})$.

Identifying the Line

Now, we check which of the given lines the point $(\frac{50}{7}, \frac{29}{7})$ lies on.

• Option 1: $y - x = 3$ $\implies \frac{29}{7} - \frac{50}{7} = -\frac{21}{7} = -3 \neq 3$.
• Option 2: $x - y = 3$ $\implies \frac{50}{7} - \frac{29}{7} = \frac{21}{7} = 3$. This matches.
• Option 3: $x + y = 11$ $\implies \frac{50}{7} + \frac{29}{7} = \frac{79}{7} \neq 11$.
• Option 4: $x + y = 12$ $\implies \frac{50}{7} + \frac{29}{7} = \frac{79}{7} \neq 12$.

The point $(a, b)$ lies on the line $x - y = 3$.