Find an equation of the plane containing the point (0,1,1) and perpendicular to the line passing through the points (2,1,0) and (1,−1,0)? 

If a plane passes through the point (x₁, y₁, z₁), the equation of a plane is represented as A(x - x₁)+B(y - y₁)+C(z - z₁) = 0.

Answer: The equation of a plane containing the point (0,1,1) and perpendicular to the line passing through the points (2,1,0) and (1,−1,0) is x - 2y + 2 = 0.

We will use the equation of a plane as A(x - x₁) + B(y - y₁) + C(z - z₁) = 0 and put the values of (x₁, y₁, z₁).

Explanation:

Let the equation of plane passing through (0,1,1) be A(x-0) + B(y−1) + C(z−1) = 0.........(i)

Now the equation for line joining points (x₁, y₁, z₁) and (x₂, y₂, z₂) is given by,

(x−x₁) / (x₂−x1) = (y−y₁) / (y₂−y₁) = (z−z₁) / (z₂−z₁)

Similarly, for points (2,1,0) and (1,−1,0) the equation of line joining is

(x−2) / (1−2) = (y−1) / (−1−1) = (z−0) / (0−0)

⇒ (x−2) / (−1) = (y−1) / (−2) = (z−0) / ...(ii)

∵ Plane (i) is perpendicular to line (ii)

A /−1 = B / −2 = C / 0 = λ(say) [∵ Normal vector of (i) is parallel to line (ii) ]

So, A = -λ, B = -2λ, C = 0λ

Put the value of A,B, and, C in equation (i).

Hence, equation of required plane is

-λ(x-0) - 2λ(y-1) + 0(z-1) = 0

⇒ x - 2y + 2 =0

Thus, The equation of the plane containing the point (0,1,1) and perpendicular to the line passing through the points (2,1,0) and (1,−1,0) is x - 2y + 2 =0.