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Let L_(1):x=y=z L_(2): x-1=y-2=z-3 be two linesLet from origin O(0,0,0) on L_(1) perpendicular is drawn to L_(2) has foot A. Segment OA is rotated about O by an angle 90^(0) such that L_(2) moves along with it. Without changing its direction & becomes L_(3) A becomes B(alpha,beta,lambda) then alpha+beta+lambda=

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Solution :LET `A(lambda+1,lambda+2,lambda+3)`
`therefore (lambda+1)+l(ambda+2)+(lambda+3)=0implieslambda=-6`
`impliesA(-5,-4,-3)`
Let LINE `L_(1) & L_(3)` LIES on plane P.
then normal to the plane P will be `5i + 4j+3k`.
implies Eq. of plane P is
`5x+4y+3z=0`
therefore Dr's of line through origin `& _|_to L_(3)`is.
`(5i+4j+3k)xx(i+j+k)`
`=lt,-2 1 gt`
therefore line OB: `(x)/(1)=(y)/(-2)=(z)/(1),`Since `OB =OA =5sqrt(2)`
`therefore` Coordinates of point B which lies on `L_(3)` is
`((5sqrt(2))/(sqrt(6)),((5sqrt(2))(-2))/(sqrt(6)),(5sqrt(2))/(sqrt(6)))` or
`((-5sqrt(2))/(sqrt(6)),((5sqrt(2))(2))/(sqrt(6)),(-5sqrt(2))/(sqrt(6)))`


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