Final answer:
The distance between the plane and the point (1, 2, 0) is 2 / √6.
Step-by-step explanation:
To find the distance between the plane and the point, we need to use the formula for the distance between a point and a plane. The formula is given by:
d = |ax + by + cz + d| / √(a^2 + b^2 + c^2),
where (a, b, c) is the normal vector to the plane, (x, y, z) is the point, and d is the constant term of the equation of the plane.
In this case, the equation of the plane is 2x - y - z = -2, so the normal vector is (2, -1, -1) and the constant term is -2.
Using the given point (1, 2, 0), we can substitute these values into the formula:
d = |2(1) + (-1)(2) + (-1)(0) + (-2)| / √(2^2 + (-1)^2 + (-1)^2)
d = |2 - 2 - 0 - 2| / √(4 + 1 + 1)
d = |-2| / √6
d = 2 / √6
So the distance between the plane and the point (1, 2, 0) is 2 / √6.