1.

Find the locus of a point,the sum of squares of whose distance from the planes `x-z=0,x-2y+z=0a n dx+y+z=0i s36`.

Answer» Given planes are
`" "x-z=0, x-2y+z=0, x+y+z=0`
Let the point whose locus is required be `P(alpha, beta, gamma)`.
According to question,
`" "(|alpha-gamma|^(2))/(2)+(|alpha-2beta+gamma|^(2))/(6)+(|alpha+beta+gamma|^(2))/(3)=36`
or `" "3(alpha^(2)+gamma^(2)-2alphagamma)+alpha^(2)+4beta^(2)+gamma^(2)-4alphabeta-4betagamma+2alphagamma`
`" "+2(alpha^(2)+beta^(2)+gamma^(2)+2alphabeta+2betagamma+2alphagamma)=36xx6`
or `" "6alpha^(2)+6beta^(2)+6gamma^(2)=36xx6`
or `" " alpha^(2)+beta^(2)+gamma^(2) = 36`
Hence, the required equation of locus is `x^(2)+y^(2)+z^(2)=36`.


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