Calculating the unit normal of a plane

Hi,

I've tried to solve a problem and I was just wondering if my answer is correct or not:

Assume three points $\displaystyle p_1, p_2$ and $\displaystyle p_3$:

$\displaystyle p_1 = (1,2,3)^T$

$\displaystyle p_2 = (2,2,1)^T$

$\displaystyle p_3 = (0,1,3)^T$

The problem is to find the unit normal of this plane.

I take the cross product of $\displaystyle p_1p_2$ and $\displaystyle p_1p_3$, I get:

$\displaystyle N = -2i + 2j - k$

I find that the length of N is $\displaystyle sqrt((-2)^2 + 2^2 + (-1)^2) = 3$.

I.e. the answer is: $\displaystyle (-2i + 2j - k) / 3$

Is this correct?