Question

 Find the equation of the plane passing through the interesection of the planes 2x + 2y -3z -7 =0 and 2x +2y - 3z -7=0 such that the intercepts made by the resulting plane on the x - axis and the z - axis are equal. 

Answer 1

 Equation of intersecting plane is 
(2x + 2y - 3z - 7) + λ (2x + 5y + 3z - 9)=0
(2 + 2λ) x+(2+5λ)y+(-3+3λ)z-7-9λ = 0              .....(1)
for intercept made on x-axsis,
put y =0 , and z=0

we get, (2+2λ)x - 7-9λ=0

`x = (7 + 9λ)/(2+2λ)`
For intercept made on z-axis
Put x = 0 and y = 0
(-3+3 λ)z-7-9 λ = 0

`z = (7+9lamda)/(-3+3lamda) `

Intercept made on x-axis = intercept made on z-axis .....(given)

`(7  + 9lamda)/(02+2lamda) = (7 + 9lamda)/(-3 +3lamda)`

`-3 + 3lamda = 2 + 2lamda`

`-3 -2 = 2lamda - 3lamda`

`-5 = -lamda`
`lamda = 5`   put in equation (1)
(2+10)x + (2+25)y+(-3+15)z - 7 - 45 =0 
12x + 27y + 12z -52 = 0 → Required equation of plane.