Calculus Homework Help Please!!?

The answer given by "la console" may well be correct.  I'm just looking for a way that feels more intuitive to me.

The area of a parallelogram is given by the magnitude of the cross product of the two different sides, so the area of each triangle in your figure is (1/2) times the magnitude of the cross product of two vectors, and the area of the whole figure is 

(1/2)(|CB "cross" CD| + |AB "cross" AD|)

= (1/2)|<1,-2,6> cross <0,3,4>| +

+ (1/2)|<-2,1,5> cross <-3,6,3>|

= (1/2)|-28i - 4j + 3k| + (1/2)|-27i - 9j - 9k|

= (1/2)[sqrt(809) + sqrt(891)].

So I get a + b = 1700.

Good luck figuring out whether "la console" is right, or whether I'm right !!!

 A (4 ; 0 ; - 1)   B (2 ; 1 ; 4)   C (1 ; 3 ; - 2)   D (1 ; 6 ; 2)

Surface area of the triangle ABC is: → a₁ = (1/2).√[det²(x) + det²(y) + det²(z)]

To calculate det(x)

_xA__xB__xC

_yA__yB__yC

__1___1___1

det(x) = xA.yC - xA.yB + xB.yA - xB.yC + xC.yB - xC.yA

det(x) = (4 * 3) - (4 * 1) + (2 * 0) - (2 * 3) + (1 * 1) - (1 * 0) = 3

To calculate det(y)

_yA__yB__yC

_zA__zB__zC

__1___1___1

det(y) = yA.zC - yA.zB + yB.zA - yB.zC + yC.zB - yC.zA

det(y) = (0 * - 2) - (0 * 4) + (1 * - 1) - (1 * - 2) + (3 * 4) - (3 * - 1) = 16

To calculate det(z)

_zA__zB__zC

_xA__xB__xC

__1___1___1

det(z) = zA.xC - zA.xB + zB.xA - zB.xC + zC.xB - zC.xA

det(z) = (- 1 * 1) - (- 1 * 2) + (4 * 4) - (4 * 1) + (- 2 * 2) - (- 2 * 4) = 17

a₁ = (1/2).√[(3)² + (16)² + (17)²] = (1/2).√554

Surface area of the triangle ADC is: → a₂ = (1/2).√[det²(x) + det²(y) + det²(z)]

To calculate det(x)

_xA__xD__xC

_yA__yD__yC

__1___1___1

det(x) = xA.yC - xA.yD + xD.yA - xD.yC + xC.yD - xC.yA

det(x) = (4 * 3) - (4 * 6) + (1 * 0) - (1 * 3) + (1 * 6) - (1 * 0) = - 9

To calculate det(y)

_yA__yD__yC

_zA__zD__zC

__1___1___1

det(y) = yA.zC - yA.zD + yD.zA - yD.zC + yC.zD - yC.zA

det(y) = (0 * - 2) - (0 * 2) + (6 * - 1) - (6 * - 2) + (3 * 2) - (3 * - 1) = 15

To calculate det(z)

_zA__zD__zC

_xA__xD__xC

__1___1___1

det(z) = zA.xC - zA.xD + zD.xA - zD.xC + zC.xD - zC.xA

det(z) = (- 1 * 1) - (- 1 * 1) + (2 * 4) - (2 * 1) + (- 2 * 1) - (- 2 * 4) = 12

a₂ = (1/2).√[81 + 225 + 144] = (1/2).√450

The area of the quadrilateral ACBD is:

= a₁ + a₂

= (1/2).√554 + (1/2).√450

= (1/2).(√554 + √450) → where: a = 554 and where: b = 450

a + b = 554 + 450 = 1004