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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! :)
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