As $\min C=5\ge \max B=5$, reduce to $G=A+B,\ H=B+C$ and apply Johnson's rule. Optimal sequence $I,IV,VII,VI,II,III,V$; total elapsed time $=59$ hrs; idle times $A=13$, $B=37$, $C=7$ hrs.

Step 1 — Check the reduction condition. For the three-machine problem, $\min A=3$, $\max B=5$, $\min C=5$. Since $\min C=5\ge\max B=5$, the problem can be converted into an equivalent two-machine problem with fictitious machines $G=A+B$ and $H=B+C$:

Step 2 — Johnson's rule on $G,H$. Pick the smallest time; if it is under $G$ schedule that job as early as possible, if under $H$ schedule it as late as possible.

• Smallest is $6$ ($V$, in $H$) $\Rightarrow$ $V$ last.
• Next $7$: $G(I)=7\Rightarrow I$ first; $H(III)=7\Rightarrow III$ next-to-last.
• Next $9$: $G(IV)=9\Rightarrow IV$ second.
• Next $10$: $G(VII)=10\Rightarrow VII$ third; then $H(VI)=10\Rightarrow VI$ toward end; $H(II)=10\Rightarrow II$ toward end.

Optimal sequence: $\;I\to IV\to VII\to VI\to II\to III\to V$.

Step 3 — Work table (In/Out on the real machines A, B, C).