Three forces in a plane keep a ring at rest, so their vectors add up to zero…

Step 1. The vector $\vec c$. $\vec c = -(1 - 2, 0 + 3, 2 + 1) = (1, -3, -3)$.

Step 2. The product $\vec b\times\vec c$. With the rows $(-2, 3, 1)$ and $(1, -3, -3)$: the $\hat i$ part is $(3)(-3) - (1)(-3) = -6$, the $\hat j$ part is $-\big[(-2)(-3) - (1)(1)\big] = -5$ and the $\hat k$ part is $(-2)(-3) - (3)(1) = 3$. So $\vec b\times\vec c = (-6, -5, 3)$.

Step 3. The product $\vec c\times\vec a$. With the rows $(1, -3, -3)$ and $(1, 0, 2)$ the result is also $(-6, -5, 3)$, so its $z$-component is $3$.

Step 4. Why. $\vec b\times\vec c = \vec b\times(-\vec a - \vec b) = -\vec b\times\vec a - \vec 0 = \vec a\times\vec b$. The same argument gives $\vec c\times\vec a = \vec a\times\vec b$ too.

Step 5. The size. $\vec a\times\vec b = (-6, -5, 3)$, so $\lvert\vec a\times\vec b\rvert = \sqrt{36 + 25 + 9} = \sqrt{70} \approx 8.37$.