Given,
a,b,c are in A.P.
⇒2b=a+c⇒a−2b+c=0
Now, on comparing above equation with ax+by+c=0 we get,
The line ax+by+c=0 passes through fixed point (1,−2)
∴P=(1,−2)
Now, we know that for infinite solution,
D=D1=D2=D3=0
D=∣1211521α3∣=0
⇒α=8
And D1=∣6β41521α3∣=0⇒β=6
So, the point Q=(α,β)=(8,6)
Hence, PQ2=72+82=49+64=113