Let P(3,2,6) be a point in space and Q be a point on the line r = (i - j + 2k) + μ(7i + j + 5k). Then the value of μ for which the vector PQ is parallel to the planex - y + 3z = 1 is:

Any point on the vector r can be taken as, Q = (1 + 7μ, (μ - 1), (5μ + 2)) gives PQ = (7μ - 2, μ - 3, 5μ - 4) Now, the PQ must be perpendicular to the normal for the given plane. 1(7μ - 2) - (μ - 3) + 3(5μ - 4) = 0 ⇒ 7μ - 2 - μ + 3 + 15μ - 12 = 0 ⇒ 21μ = 11 ⇒ μ = 11/21 This value of μ is not in the options. Let's check the calculation again. 1(7μ - 2) - 1(μ - 1) + 3(5μ - 4) = 0 7μ - 2 - μ + 1 + 15μ - 12 = 0 21μ = 13 μ = 13/21 This value of μ is also not in the options. Let's assume that PQ is parallel to the plane. Then the dot product of PQ and the normal vector of the plane should be 0. Normal vector of the plane x - y + 3z = 1 is (1, -1, 3). PQ = (7μ - 2, μ - 3, 5μ - 4) (7μ - 2, μ - 3, 5μ - 4) . (1, -1, 3) = 0 7μ - 2 - (μ - 3) + 3(5μ - 4) = 0 7μ - 2 - μ + 3 + 15μ - 12 = 0 21μ = 11 μ = 11/21 This is still not in the options. Let's reconsider the problem statement. Let's use the condition that PQ is parallel to the plane x - y + 3z = 1. The normal vector to the plane is N = (1, -1, 3). The vector PQ is given by Q - P = (1 + 7μ - 3, μ - 1 - 2, 5μ + 2 - 6) = (7μ - 2, μ - 3, 5μ - 4). For PQ to be parallel to the plane, the dot product of PQ and N must be zero: (7μ - 2, μ - 3, 5μ - 4) . (1, -1, 3) = 0 7μ - 2 - (μ - 3) + 3(5μ - 4) = 0 7μ - 2 - μ + 3 + 15μ - 12 = 0 21μ = 11 μ = 11/21 Since this value is not among the options, there might be an error in the problem statement or the options provided. However, if we assume there was a mistake and we are looking for a value of μ such that the dot product is zero: (7μ - 2) - (μ - 3) + 3(5μ - 4) = 0 21μ - 11 = 0 μ = 11/21 This is not in options. There must be an error in the question or options.