Question
Find the equation of the plane which contains the line of intersection of the planes 
and which is perpendicular to the plane 
Solution
The equations of given planes are
and
and
Any plane through the intersection of planes (1) and (2) is
(x + 2 y + 3 z – 4) + k (2 x + y – z + 5) = 0 ...(3)
i.e. (2 k + 1) x + (k + 2) y + (– k + 3) z + (5 k – 4) = 0
Direction ratios of the its normal are 2 k + 1, k + 2, – k + 3.
Again consider the plane
or 
or 
...(4)
Direction ratios of its normal are 5, 3, -6
Since plane (3) is perpendicular to plane (4)
∴ 5 (2 k + 1) + 3 (k + 2) + (–6) (– k + 3) = 0
∴ 10 k + 5 + 3 k + 6 + 6 k – 18 = 0
Putting 
or 19 (x + 2y + 3z – 4) + 7 (2 x + y – z + 5) = 0
or 19x+ 38y + 57z – 76 + 14x + 7y – 7z + 35 = 0
or 33x + 45y + 50z – 41 = 0
which is required equation of plane.
