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  • The distance between Point (-1 ,-5,-10 ) and the point of intersection of Line x-2 / 3=y+1 / 4 =z-2 / 12 with the plane x-y+z=5 .

The distance between Point (-1 ,-5,-10 ) and the point of intersection of Line x-2 / 3=y+1 / 4 =z-2 / 12 with the plane x-y+z=5 .

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Solution: We have , 

                 L_1:x-2/3=y+1/4=z-2/12 , and   P:x-y+z=5

       Now ,   L_1:x-2/3=y+1/4=z-2/12=r  

       General point on the Given Line L_1   ,A(3r+2,4r-1,12r+2)

        ecause     Point A Lies on thr plane P 

         herefore           (3r+2)-(4r-1)+(12r+2)=5  Rightarrow r=0

      Point  A(2,-1,2)  ,Hence distance between A(2,-1,2) and B(-1,-5,-10)

        Rightarrow         AB=sqrt9+36+144=sqrt189   

Posted by

Deependra Verma

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