The correct option is A (3,2)
Let f(x,y)=x2−6xy+y2+6x+14y−2=0
Partially differentiate w.r.t. x and equate to zero
δfδx=2x−6y⋅1+0+6+0+0=0 (treat y as constant)
⇒x−3y+3=0 ⋯(1)
Again, partially differentiate w.r.t. y and equate to zero
δfδy=−6x+2y+14=0 (treat x as constant)
⇒y−3x+7=0 ⋯(2)
Solving (1) and (2),
x=3 and y=2
∴ Centre of given conic is (3,2)