Question 368274
a) Use the cross product to find a vector perpendicular to both vector and the plane.
{{{V=AxB}}}
V=(6,0,1)x(2,1,0)
.
.
.
b) Again use the cross product
A:(1,2,3)
B:(4,-3,2)
C:(8,1,1)

{{{Area=(1/2)(ABxAC)}}}
where AB is the vector from A to B and AC is the vector from A to C.
Check section 4.1 of http://en.wikipedia.org/wiki/Triangle for a derivation of this method.