The direction cosines of any normal to the xy plane are

The correct option is 1, 1, 0.

Let us understand the step-by-step explanation of this result in the context of three-dimensional coordinate geometry:
1. A plane in three-dimensional space is represented by the general equation:
$$ax+by+cz+d=0$$ where the coefficients $a$, $b$, and $c$ represent the direction ratios of the normal to the plane.

2. For the given correct option 1, 1, 0, these values correspond to the direction ratios of the normal vector.
To find the direction cosines $(l,m,n)$, we divide each coordinate by the magnitude of the vector:
$$\text{Magnitude}=\sqrt{1^2+1^2+0^2}=\sqrt{2}$$ Thus, the corresponding normalized direction cosines would be:
$$(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0)$$ This shows that the normal vector lies entirely within the xy-plane projection (having a z-component of 0), matching the proportion of the direction components given in the correct option 1, 1, 0.