In `R_(3) , ` consider the planes `P_(1):y=0and P_(2) : x+z=1. Let P_(3)` be a plane , different from `P_(1) and P_(2)` , which passes through the interesection of `P_(1) and P_(2) `I fhte distance of the (0,1,0) from `p_(3)` is 1 and the distance of a point `(alpha , beta , gamma )` from `P_(3)` is 2, then which of the following relations is ( are ) true ?

Options

A. `2alpha + beta + 2gamma +2 = 0 `
B. `2alpha -beta + 2gamma +4=0`
C. `2alpha + beta - 2gamma- 10 =0`
D. `2alpha- beta+ 2gamma-8=0`

Answer

b, d

Explanation

Clearly, plane `P_(3)` is `P_(2)+ lamdaP_(1)=0`.
`rArr" "x+lamday+z-1=0`
Distance of this plane from point `(0, 1, 0)` is 1.
`rArr " "(0+lamda+0-1)/(sqrt(1+lamda^(2)+1))= pm 1`
`therefore " "lamda= - (1)/(2)`
Thus, equation of `P_(3)` is `2x-y+2z-2=0`.
Distance of this plane from point `(alpha, beta, gamma)` is 2.
`rArr" "|(2alpha-beta+2gamma-2)/(3)|=2`
`rArr" "2alpha-beta+2gamma= 2pm 6`
Thus, option (b), (d) are correct.