the locus of the point,the sum of the squares of whose distance from the planes x+y+z=0,x-y=0,x+y-2z=0 is 5 is

let the point be (h,k,l)

distance from a plane ax+by+cz+d is

(ah+bk+cl)/(√a²+b²+c²)

so distance from first plane is,

(h+j+k)/√3.....................(1)

distance from second plane is,

(h-k)/√2.....................(2)

distance from third plane is,

(h+k-2l)/√6.....................(3)

sum of squares is 5

so the sum of squares of (1), (2) and (3) is the required locus! :)