The correct option is A (2,−1)
Given equation is
x(a+2b)+y(a+3b)=a+b⇒a(x+y−1)+b(2x+3y−1)=0⋯(1)
λ1L1+λ2L2=0 represents family of straight lines which will always pass through the point of intersection of lines L1=0,L2=0
i.e., x+y−1=0 and 2x+3y−1=0,
Solving above equations, we get
⇒y=−1,x=2
Hence, the fixed point is (2,−1).