**Find the area of a triangle with vertices A(1, 3, 5), B(-2, -3, -4) and C(0, 3, -1)**

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You need to use the following formula to evaluate the area of triangle `Delta ABC` , using the vertices, such that:

`A_(Delta ABC) = (1/2) |bar (AB) x bar (AC)|`

You need to evaluate the cross product, such that:

`|bar (AB) x bar (AC)| = |[(bar i, bar j, bar k),(x_B - x_A, y_B - y_A, z_B - z_A), (x_C - x_A, y_C - y_A, z_C - z_A)]|`

`|bar (AB) x bar (AC)| = |[(bar i, bar j, bar k),(-2 - 1, -3 - 3, -4 - 5), (0-1, 3-3, -1-5)]|`

`|bar (AB) x bar (AC)| = |[(bar i, bar j, bar k),(-3, -6, -9), (-1, 0, -6)]|`

`|bar (AB) x bar (AC)| = |36 bar i + 9bar j - 6 bar k - 18 bar j|`

`|bar (AB) x bar (AC)| = |36 bar i - 9 bar j - 6 bar k|`

`|bar (AB) x bar (AC)| = sqrt(36^2 + 81 + 36)`

`|bar (AB) x bar (AC)| ~~ 37.58 `

**Hence, evaluating the area of the given triangle Delta ABC, yields **`A_(Delta ABC) = 18.79.`

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