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6x+4y+2z=12
12 12 12
/6 /4 /2
-------------
2 3 6 are our intercepts

one way is we can take these and form vectors to get an angle to do a sin area formula with

3.164 squre unit approx using area = squrt{ s(s-a)(s-b)(s-c)) ;s=(a+b+c)/2

my error is in the sqrt(58) :)
sqrt(4+9) * sqrt(9+36)
sqrt(13*45)
3 sqrt(65)

PLEASE SEE THE DRAWING........

i got a few more errors, but the concepts seems solid ;)

So 2,3,6 are the intercepts,
sqrt(13), sqrt(45), sqrt(40) are the sides.

right, and you can heron that for sheer simplicity

Some "exact" answers would be nice! :)

Agree, we also want as many different ways as possible.

|dw:1324567464017:dw|

sqrt(45) sin(y) = height
sqrt(13) = base

b*h/2 = 3sqrt(14) if i did it right that time

the bh/2 is after the vectors give us a sin for an angle to determine the height

work in progress here
|dw:1324568575430:dw|
\[\int_{0}^{6}distance.x.to.y\ dz\]

hmm, wolf says that about 19, which is a bit off from the 3sqrt(14)
i wonder why

348 not 468 ...
but still thats off

oh well, it was a thought. guess i dont know how to freehand integrals yet :)