(a) $PA^{2}=4PB^{2}$ yields $3x^{2}+3y^{2}-4x-14y+15=0$. (b) The critical path is $0\text{-}1\text{-}3\text{-}6\text{-}7$ with completion time $31$ weeks.

Part (a) — Locus of the point.

Let the moving point be $P(x,y)$, with $A(2,1)$ and $B(1,2)$. Given $PA:PB=2:1$, so $PA=2\,PB\Rightarrow PA^{2}=4\,PB^{2}$.

Step 1 — Write the squared distances.

$(x-2)^{2}+(y-1)^{2}=4\left[(x-1)^{2}+(y-2)^{2}\right].$

Step 2 — Expand.

$x^{2}+y^{2}-4x-2y+5=4\left[x^{2}+y^{2}-2x-4y+5\right]=4x^{2}+4y^{2}-8x-16y+20.$

Step 3 — Bring all terms to one side.

$0=3x^{2}+3y^{2}-4x-14y+15.$

So the locus is $3x^{2}+3y^{2}-4x-14y+15=0$.

Part (b) — Network analysis (CPM).

Compute earliest event times $E$ (forward pass) and latest event times $L$ (backward pass) with durations 0-1:3, 1-2:8, 1-3:12, 2-4:6, 2-5:3, 3-4:3, 3-6:8, 4-7:5, 5-7:3, 6-7:8.

Forward pass (earliest event times):

$E_0=0,\ E_1=3,\ E_2=11,\ E_3=15,\ E_4=\max(11+6,15+3)=18,\ E_5=14,\ E_6=15+8=23,$

$E_7=\max(18+5,14+3,23+8)=31.$

Backward pass (latest event times):

$L_7=31,\ L_6=23,\ L_5=28,\ L_4=26,\ L_3=\min(26-3,23-8)=15,\ L_2=\min(26-6,28-3)=20,\ L_1=\min(20-8,15-12)=3,\ L_0=0.$

Activity table (EST $=E_i$, EFT $=E_i+t$, LFT $=L_j$, LST $=L_j-t$):