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).
