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**eXist** For the first problem, you need to find a normal vector that defines the plane. To do this find two vectors in the plane using those points and cross them:

$\displaystyle A = (5, 1, 2) - (-1, 3, 0) = <6, -2, 0>$

$\displaystyle B = (4, -3, 1) - (-1, 3, 0) = <5, -6, 0>$

$\displaystyle A X B = <0, 0, -26> = C$

Now this is the normal vector that defines the plane, we need to get the unit vector:

Unit Vector: $\displaystyle \frac{C}{|C|} = \frac{<0, 0, -26>}{26} = <0, 0, -1>$ also $\displaystyle <0, 0, 1>$

We must include both the negative and positive of this unit vector because it is perpendicular either way. I hope that's pretty clear, if not, I'll explain more just ask.

For the second problem, take the absolute value of the determinant divided by 2:

$\displaystyle \left(\begin{array}{cccc}3&2&1\\2&4&6\\-1&2&5\end{array}\right) = 0$

I believe these points to be co-linear since I calculated the determinant to be 0, but I'm going to check again. So for now, I believe the area of the triangle to be 0.