The curl of F⃗ =(z−x−y)i⃗ +(x−y−z)j⃗ +(y−z−x)k⃗ 𝐹→=(𝑧−𝑥−𝑦)𝑖→+(𝑥−𝑦−𝑧)𝑗→+(𝑦−𝑧−𝑥)𝑘→ is

The curl of a vector field F is given by the cross product of the del operator and the vector field. In this case, the vector field F is given by F = (z - x - y)i + (x - y - z)j + (y - z - x)k.

The del operator is given by ∇ = ∂/∂x i + ∂/∂y j + ∂/∂z k.

So, the curl of F, denoted by ∇ x F, is given by the determinant of the following 3x3 matrix:

| i j k | | ∂/∂x ∂/∂y ∂/∂z | | z-x-y x-y-z y-z-x |

Calculating this determinant gives:

i(∂(y-z-x)/∂y - ∂(x-y-z)/∂z) - j(∂(y-z-x)/∂x - ∂(z-x-y)/∂z) + k(∂(x-y-z)/∂x - ∂(z-x-y)/∂y)

= i(1 - (-1)) - j(1 - (-1)) + k(1 - 1)

= 2i - 2j + 0k

So, the curl of F is 2i - 2j.