Edit: DONE!!!I updated question so you have answer. Hope it helps somebody. <3
A four-sided pyramid $ABCDE$ whose base is the parallelogram ABCD is given. If $A(2,3,1)$, $B(4,1,-2)$, $C(6,3,7)$, $E(-5,-4,8)$ calculate the volume of the pyramid and the height drawn from the top $E$.
I choosed vectors BA=(-2,2,3), BC=(2,2,9) and BE=(-9,-5,10). I know that volume of parallelepiped is (BA x BC)*BE. I found determinant of this and it's -308 but i'm using |-308|=308.
This four-sided pyramid contains 2 tetraedars. Formula for volume of tetraedars is $(1/6)(BA x BC)*BE$, so 308*(1/6). Multiply this with 2 so you get: $V = (1/3)*308$ and thats volume of pyramid. Next is area of paralellogram which is base of pyramid. $P = |BA x BC|$ =(12,24,-8) = 28. V = (1/3)BH so H = 11