BITSAT 2018 Slot 2 Solved Paper

Let distances of a point p(x,y,z) from the planes x+y+z=0,x−z=0 and x−2y+z=0 are

x+y+z

√3

,

x−z

√2

and

x−2y+z

√6

respectively, then the sum of the squares of distances, is as
⇒(

x+y+z

√3

)2+(

x−z

√2

)2 +(

x−2y+z

√6

)2=p2
⇒2(x+y+z)2+3(x−z)2+(x−2y+z)2 =6p2
⇒2x2+2y2+2z2+4xy+4yz +4zx+3x2+3z2−6xz+x2+4y2+z2−4xy −4yz+2xz=6p2
⇒6x2+6y+6z2=6p2
⇒x2+y2+z2=p2
Hence, option (d) is correct.