Find the distance of the point (x, y, z) from the point of intersection of the line r = 2i - j + 2k + λ(3i + 4j + 2k) and the plane r.(i - j + k) = 5.

The plane can be written as x - y + z = 5. Any point on the line is (2 + 3λ, -1 + 4λ, 2 + 2λ). Now, for the point of intersection, substitute the equation of the point in the plane to obtain parameter λ.
∴ 2 + 3λ - (-1 + 4λ) + 2 + 2λ = 5
∴ 5 + λ = 5
∴ λ = 0
Hence, the point of intersection of the line and plane is (2, -1, 2).
Distance between (x, y, z) and (2, -1, 2) = √((2 + 1)² + (-1 + 5)² + (2 + 10)²) = √(9 + 16 + 144) = √169 = 13
This is the required answer.