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Treat each column as the payoff from one unit of initial investment, so the cost of holdings $h=(h_1,h_2,h_3{)}^t$ is $h_1+h_2+h_3$. The first asset weakly dominates the third at the same price, and so does the second. Long the first and short the third gives zero initial cost and terminal payoff $(0,0,2{)}^t$. Long the second and short the third gives payoff $(2,0,0{)}^t$, also at zero cost. These are arbitrages provided the displayed states have positive probability. Combining them gives the zero-cost position $(1,1,-2{)}^t$ with payoff $(2,0,2{)}^t$.

More generally, imposing zero cost means $h_3=-h_1-h_2$, and its payoff is $(2h_2,0,2h_1{)}^t$. Thus every $h_1,h_2\ge 0$ with at least one positive gives a pure-investment arbitrage. The market is not arbitrage-free.

There is nevertheless no arbitrage of the other requested kind, with every future net cash flow identically zero. Indeed the middle-state payoff is $2(h_1+h_2+h_3)$, so a zero terminal payoff already forces zero initial cost. The other two states then force $h_1=h_2=0$, hence $h_3=0$. Equivalently, the return matrix has determinant $8$ and is invertible. ThereforeThe unique vector pricing the three unit-investment payoffs is $(0,\frac{1}{2},0{)}^t$. It is nonnegative but not strictly positive. This illustrates why nonnegative state prices need not exclude arbitrage: strictly positive state prices would assign positive cost to either of the nonzero nonnegative payoffs above.

For holdings $h\in {\mathbb{R}}^n$, let $V_0=h\cdot P_0$ and $V_1=h\cdot P_1$. Under the investment-consumption arbitrage convention used here, an arbitrage satisfiesThe initial consumption is $-V_0$, and the terminal consumption is $V_1$: there is a possible gain with no external funding and no negative consumption. A pure-investment arbitrage additionally requires $V_0=0$. This convention is important because the last part distinguishes initial consumption from pure investment.

Write $(a,b,c)$ for holdings of cash, the stock and the European put option, and let $\gamma =-V_0$ be initial consumption. At the given price, $a=-10b-c-\gamma$, and the three terminal payoffs, in descending order of the stock price, areNonnegative terminal payoffs require $c+\gamma \le 2b$ and $c-\gamma \ge b$. Together with $\gamma \ge 0$, these imply $b\ge 2\gamma \ge 0$. If $b=0$ they force $\gamma =c=a=0$, which is not an arbitrage. If $b>0$, the highest-state payoff is at least $3b>0$, so every such choice is an arbitrage. Consequently the complete set isThe pure-investment arbitrages are exactly the choices $\gamma =0$:For example, $(a,b,c)=(-11,1,1)$ costs zero and pays $(4,1,0)$.