The LPP maximize $z = 2x + 5y$ subject to $x + 3y \leq 6$, $2x + 6y \leq 18$, $x \geq 0$, $y \geq 0$ has

The given constraints are:
$x + 3y \leq 6$
$2x + 6y \leq 18 \Rightarrow x + 3y \leq 9$
$x \geq 0, y \geq 0$

The second constraint $x + 3y \leq 9$ is redundant since $x + 3y \leq 6$ is a stricter condition. The feasible region is bounded by $x + 3y \leq 6$ in the first quadrant, which forms a triangle.

The vertices of the feasible region are $(0,0)$, $(6,0)$, and $(0,2)$.

Evaluating the objective function $z = 2x + 5y$ at these vertices:
At $(0,0)$, $z = 2(0) + 5(0) = 0$
At $(6,0)$, $z = 2(6) + 5(0) = 12$
At $(0,2)$, $z = 2(0) + 5(2) = 10$

The maximum value of $z$ is $12$, which occurs exactly at one vertex $(6,0)$.

Therefore, the LPP has a unique solution.

Answer: Unique solution