Question

A plane passes through (2, 1, 2) and (1, 2, 1) and parallel to line 2x = 3y and Z = 1 then plane also passes through the point?
(-6, 2, 0)
(6, -2, 0)
(-2, 0, 1)
(2, 0, 1) ​

Answer 1

A plane passes through (2, 1, 2) and (1, 2, 1) and parallel to line 2x = 3y and Z = 1

To find : The plane also passes through the point .....

solution : equation of plane passes through (2, 1, 2) is given by,

a(x - 2) + b(y - 1) + c(z - 2) = 0 ......(1)

as the same plane also passes through (1, 2, 1)

so, a(1 - 2) + b(2 - 1) + c(1 - 2) = 0

⇒-a + b - c = 0

⇒a - b + c = 0 .......(2)

line 2x = 3y and z = 1, is parallel to plane (1)

x/3 = y/2 = (z - 1)/0, is parallel to plane (1),

so, 3a + 2b + 0c = 0

⇒3a + 2b = 0 ......(3)

from cramer's law,

a/(-1 × 0 - 1 × 2) = -b/(1 × 0 - 1 × 3) = c/(1 × 2 - (-1) × 3)

a/-2 = -b/-3 = c/5

a/-2 = b/3 = c/5

so , plane is -2(x - 2) + 3(y - 1) + 5(z - 2) = 0

⇒-2x + 3y + 5z + 4 - 3 - 10 = 0

⇒-2x + 3y + 5z - 9 = 0 now you can check all given points

here we get, (-2, 0, 1) satisfies the plane.

so, plane also passes through the point (-2, 0, 1)