Evaluate using Stokes' theorem

Stokes' theorem: ∮_C F·dr = ∬_S (curl F)·n dS, with F = (x + y) i + (2x − z) j + (y + z) k.

The triangle with vertices A(2,0,0), B(0,3,0), C(0,0,6) lies on the plane x/2 + y/3 + z/6 = 1, that is 3x + 2y + z = 6.

curl F = (∂R/∂y − ∂Q/∂z) i + (∂P/∂z − ∂R/∂x) j + (∂Q/∂x − ∂P/∂y) k
       = (1 + 1) i + (0 − 0) j + (2 − 1) k = 2 i + k

n = (3i + 2j + k)/√14        (upward, for A → B → C anticlockwise seen from above)
(curl F)·n = (6 + 0 + 1)/√14 = 7/√14
dS = dx dy/|n·k| = √14 dx dy

So (curl F)·n dS = 7 dx dy. The projection on the xy-plane is the triangle with vertices (0,0), (2,0), (0,3), of area ½ · 2 · 3 = 3.

∮_C F·dr = 7 × 3 = 21

Check (direct): summing the line integrals along A→B, B→C, C→A gives 21 as well.

Answer: 21 (for the orientation A → B → C → A).