For a given Linear Programming problem, the objective function is $Z=3 x+2 y$ Subject to constraints are $\begin{aligned} & 4 x+3 y \leq 60 \\ & x \geq 3 \\ & y \leq 2 x \\ & y \geq 0 \end{aligned}$ P is one of the corner points of the feasible region for the given Linear Programming problem. Then the coordinate of P is

The feasible region is defined by the constraints $4x + 3y \leq 60$, $x \geq 3$, $y \leq 2x$, and $y \geq 0$.

First, identify the intersection points of the boundary lines:
1. The line $x = 3$ intersects $y = 0$ at $(3, 0)$.
2. The line $x = 3$ intersects $y = 2x$ at $(3, 6)$.
3. The line $y = 2x$ intersects $4x + 3y = 60$. Substituting $y = 2x$ into the equation: $4x + 3(2x) = 60 \Rightarrow 4x + 6x = 60 \Rightarrow 10x = 60 \Rightarrow x = 6$. Then $y = 2(6) = 12$. So, the point is $(6, 12)$.
4. The line $4x + 3y = 60$ intersects $y = 0$ at $4x = 60 \Rightarrow x = 15$. So, the point is $(15, 0)$.

The corner points of the feasible region are $(3, 0)$, $(3, 6)$, $(6, 12)$, and $(15, 0)$.

Comparing these with the given options, the point $(3, 6)$ is a corner point of the feasible region.

Answer: $(3,6)$