Given that required plane passes through points P(2, 4, 2) and Q(2, 3, 5) and parallel to a line whose direction ratios are 3, -2, 1.

Now, direction ratios of line AB are 0, -1, 3

Let the direction ratios of normal to the plane are A, B and C.

Since, point P and Q lie on plane, therefore line joining points P and Q is perpendicular to normal.

∴ 0 x A + (-1) x B + 3 x C = 0 (condition of two perpendicular lines)

⇒ -B+3C = 0

⇒ B=3C .....(i)

Since, given plane is parallel to line 3i - 2j + k

∴ Normal to plane is perpendicular o line 3i - 2j + k

∴ 3A-2B+C = 0

⇒ 3A-6C+C = 0 (from(i))

⇒ 3A = 5C

⇒ A = 5C/3 .....(ii)

Hence, the direction ratios of normal to required plane are 5C/3, 3C, C or 5, 9, 3. (By putting C=3)

Therefore, the equation of the plane is 5x+9y+3z = d, where d is perpendicular distance of origin from plane.

Since, point (2, 4, 2) lies on plane.

∴ d = 2 x 5 + 4 x 9 + 2 x 3

= 10+36+6

= 52

∴ Equation of required plane is 5x+9y+3z = 52.

Equation of line joining the points 2i+j+3k and 4i+2j+3k is r = a+λ(b-a)

⇒ r = (2i+j+3k) + λ((4-2)i + (2-1)j + (3-3)k

⇒ xi+yj+zk = (2+2λ)i + (1+λ)j + 3k (∵ r = xi+yj+zk)

⇒ x = 2+2λ, y = 1+λ, z = 3

Hence, locus of arbitrary point on line joining points (2, 1, 3) and (4, 2, 3) is given by.

(x, y, z) = (2(1+2λ)), 1+λ, 3j

Since, line meets the plane, therefore equation of arbitrary point on line satisfies equation of plane.

∴ 5 x 2(1+λ) + 9 x (1+λ) + 3 x 3 = 52

⇒ 10+10λ+9+9λ+9 = 52

⇒ 19λ = 52-28 = 24

⇒ λ = 24/19

∴ point which lies on both line and plane is (x, y, z) = (86/19, 43/19, 3).

Hence, plane meets the line at (86/19, 43/19, 3).