The corner points of the feasible region of the LPP:

Minimize Z = −50x + 20y subject to 2x − y ≥ −5, 3x + y ≥ 3, 2x − 3y ≤ 12, and x, y ≥ 0 are:

The correct option is (0, 3), (0, 5), (1, 0), (6, 0).

To find the corner points of the feasible region, we must analyze the given system of linear inequalities and identify the region that satisfies all constraints simultaneously.

The given objective function and constraints are:
Minimize $Z=-50x+20y$
subject to the constraints:
1. $2x-y\ge -5$
2. $3x+y\ge 3$
3. $2x-3y\le 12$
4. $x\ge 0,y\ge 0$ (which restricts the feasible region to the first quadrant).

Let us construct the corresponding boundary lines and find their intercepts on the coordinate axes:

Line 1: $2x-y=-5$
- If $x=0$, then $y=5$. This gives the point $(0,5)$.
- If $y=0$, then $x=-2.5$. This gives the point $(-2.5,0)$.

Line 2: $3x+y=3$
- If $x=0$, then $y=3$. This gives the point $(0,3)$.
- If $y=0$, then $x=1$. This gives the point $(1,0)$.

Line 3: $2x-3y=12$
- If $x=0$, then $y=-4$. This gives the point $(0,-4)$.
- If $y=0$, then $x=6$. This gives the point $(6,0)$.

Now, we identify the corner points of the feasible region by checking the intersections of these boundary lines in the first quadrant:
- The y-axis ($x=0$) bounds the region between the points $(0,3)$ and $(0,5)$.
- The x-axis ($y=0$) bounds the region between the points $(1,0)$ and $(6,0)$.
- Thus, the vertices enclosing the feasible region in the first quadrant are:
$(0,3)$, $(0,5)$, $(1,0)$, and $(6,0)$.