Fixed OA. Indeed there was a mismatch when I checked carefully

A and D are correct answer choices

Thank you sir to pointing out

Since there are n sides in the polygon, the number of vertices is also n. We need to find the number of diagonals that can be formed using the n vertices. Number of ways in which any 2 vertices can be selected to form a line

$C^n_2=\frac{n(n-1)}{2}$

Among the above number of lines, we have the n sides included; we must exclude them. Thus, the number of diagonals = Total number of lines– Number of sides

$\frac{n(n-1)}{2}-n$$=\frac{n(n-3)}{2}$

We know that the difference between the number of diagonals and the number of sides is 3.

Thus, we have the following 2 cases:

(A) Number of sides– Number of diagonals = 3

$\frac{n-n(n-3)}{2}=3$

Expand and we have

$n^2-5n+6=0$

Find the roots

N=2 or N=3

Since, to form a polygon, we need a minimum of 3 sides, the only value of n 3


(B) Number of diagonals–Number of sides=3


$\frac{n(n-3)}{2}-n=3$

Expand and find the roots

$(n-6)(n+1)=0$

N=6 or N=-1

Since the number of sides must be positive, the only value of n = 6

Thus, the possible values of n are 3 OR 6.

Alternately, we can simply use the values of n given in the options to calculate the number of
diagonals and check whether the difference is 3.

For n = 3, the number of diagonals is 0 (a triangle has no diagonal) and hence, the difference is 3.

For n = 6, a hexagon, the number of diagonals = 3, and hence, the difference is also 3.

The correct answers are options A and D.