`(P)////(Q)=>vec(n_((P)))=vec(n_((Q)))=(2;-1;2)`

`=>` `(Q): 2x-y+2z+D=0`

Mà `A in (Q)=> 2*3-2+2*(-1)+D=0`

`=>` `D=-2`

`=>` `(Q): 2x-y+2z-2=0`

Ta có: `vec(n_((R)))=[vec(n_((Q))),vec(n_((Oxy)))]=(-1;-2;0)`

`=>` `(R): -1*x-2y+d=0`

Mà `d(B, (R))=sqrt5`

`=>` `(|-3+2*1-d|)/(sqrt(1^2+2^2+0^2))=sqrt5`

`=>` `|d+1|=5`

`=>` `[(d+1=5),(d+1=-5):}`

`=>` `[(d=4),(d=-6):}`

Với `d=4=> (R): -x-2y+4=0<=> x+2y-4=0` (thỏa mãn)

Với `d=-6=> (R): -x-2y-6=0<=> x+2y+6=0` (loại)

`=>` `(R): x+2y-4=0`

`=>` `{(b=2),(d=-4):}`

`=>` `b+d=2+(-4)=-2`