Find the distance of the point (1, 5, 10) from the point of intersection of the line →r = 2î - ĵ + 2k + λ(3î + 4ĵ + 2k) and the plane →r · (î - ĵ + k) = 5.

The line is x - 2/3 = y + 1/4 = z - 2/2 = λ
Any point on it is P(2 + 3λ, -1 + 4λ, 2 + 2λ)
P lies on x - y + z = 5
Therefore, 2 + 3λ - (-1 + 4λ) + 2 + 2λ = 5
i.e., λ = 0
The point of intersection is : P(2, -1, 2)
The required distance from point Q(1, 5, 10) = √(1 - 2)² + (5 + 1)² + (10 - 2)² = √1 + 36 + 64 = √101 units.