We are given two points on the plane: A(1,0,0) and B(0,1,0). To find the normal to the plane, we need to find the cross product of the vectors AB and the vector that makes an angle of π/3 with the normal.

First, let's find AB:

AB = B - A = (0,1,0) - (1,0,0) = (-1,1,0)

Next, let's find the vector that makes an angle of π/3 with the normal. Let this vector be v. We know that the dot product of two vectors is equal to the product of their magnitudes times the cosine of the angle between them. So we can write:

v · n = |v| |n| cos(π/3)

Since we want v to be a unit vector, we can set |v| = 1. We also know that the normal to the plane is perpendicular to AB, so we can set n = AB × k, where k is some constant. Then we have:

v · (AB × k) = |AB × k| cos(π/3)

k v · (AB × k) = |AB × k| cos(π/3)

Since AB and k are both perpendicular to the normal, AB × k is parallel to the normal, so we can write:

v · (normal) = |normal| cos(π/3)

Substituting in the values we know:

k v · (-1,1,0) = |(-1,1,0)| cos(π/3)

k v · (-1,1,0) = √2/2

We can choose k to be any non-zero constant, so let's choose k = 1. Then we have:

v · (-1,1,0) = √2/2

v(-1) + v(1) = √2/2

v = √2/4

So the vector that makes an angle of π/3 with the normal is v(-1,1,0) = (-√2/4, √2/4, 0).

Finally, we can find the normal to the plane by taking the cross product of AB and the vector we just found:

normal = AB × v = (-1,1,0) × (-√2/4, √2/4, 0) = (-√2/4, -√2/4, 1/2)

So the equation of the plane is:

-√2/4(x - 1) - √2/4(y - 0) + 1/2(z - 0) = 0

Simplifying:

-√2/4 x - √2/4 y + 1/2 z - √2/4 = 0

Multiplying by -4/√2 to get rid of the fraction:

-2√2 x - 2√2 y + 2√2 z - 2 = 0

Simplifying:

x + y - z = 1/√2

So the equation of the plane is x + y - z = 1/√2, and the normal to the plane is (-√2