Given – A plane passes through points (1, 2, 3) and (0, - 1, 0) and is parallel to the line

To find – Equation of the plane

Tip – If a plane passes through points (a’, b’, c’), then its equation may be given as a(x - a’) + b(y - b’) + c(z - c’) = 0

Taking points (1, 2, 3):

a(x - 1) + b(y - 2) + c(z - 3) = 0…………(i)

The plane passes through (0, - 1, 0):

a(0 - 1) + b(- 1 - 2) + c(0 - 3) = 0

i.e. a + 3b + 3c = 0………………(ii)

The plane is parallel to the line

Tip – The normal of the plane will be normal to the given line since both the line and plane are parallel.

Direction ratios of the line is (2, 3, - 3)

Direction ratios of the normal of the plane is (a, b, c)

So, 2a + 3b - 3c = 0………………………(iii)

Solving equations (ii) and (iii),

[α → arbitrary constant]

Putting these values in equation (i) we get,