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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.