Find coordinates of two vector product given each vector in coordinate format $\vec{a} = (-1,0,2), \vec{b} = (3,-1,1), \vec{c} = (2,1,2) $

I need to find coordinates of vectors, as well as modulus of the following:

1) $\vec{a}$ x $\vec{b}$

2) $\vec{b}$ x $\vec{c}$

I am not sure I not understand how to find above.

Should I sum up coordinates to get $\times$? And afterwards take squares to find modulus?
Using formulas given in link by @Rory Daulton



Thus 
$\vec{a}$ x $\vec{b} = (0+2)i + (6+1)j + (1+0)k = 2i + 7j + 1k $

If you do not remember formula, it could be easier to remember determinant form:



from where determinant can be calculated using two different formulas:

 

or



Thus $\vec{a}$ x $\vec{b} = (0\times{i} + 6\times{j} + 1\times{k}) - (-2\times{i} - 1\times{j} + 0\times{k})=2{i} + 7{j} + 1{k}$

Modulus can be found as $\sqrt{2\times2 + 7\times7 + 1\times1}=\sqrt{54}$

The very same idea is for $\vec{b}$ x $\vec{c}$, finding
$\vec{a}$ x $\vec{b} = -3i-4j+5k$ and modulus = $\sqrt{42}$
Using formulas given in link by @Rory Daulton



Thus 
$\vec{a}$ x $\vec{b} = (0+2)i + (6+1)j + (1+0)k = 2i + 7j + 1k $

If you do not remember formula, it could be easier to remember determinant form:



from where determinant can be calculated using two different formulas:

 

or



Thus $\vec{a}$ x $\vec{b} = (0\times{i} + 6\times{j} + 1\times{k}) - (-2\times{i} - 1\times{j} + 0\times{k})=2{i} + 7{j} + 1{k}$

Modulus can be found as $\sqrt{2\times2 + 7\times7 + 1\times1}=\sqrt{54}$

The very same idea is for $\vec{b}$ x $\vec{c}$, finding
$\vec{a}$ x $\vec{b} = -3i-4j+5k$ and modulus = $\sqrt{42}$