The symmetric form of the equation of the line of intersection of the planes x+y-z = 1 and 2x-3y+z=2 is

Dear deepika

planes x+y-z = 1 and 2x-3y+z=2

let the direction ratio of the line are a,b,c

so a+b-c =0   and 2a-3b+c =0

so from these two equation

 a/2  = b/3 =c/5

now we have to find out the point of intersection of these two planes.

let x=0

so  y-z =1   and   -3y + z =2

 so y=-3/2

   z = -5/2

so equation of line become

 (x-0)/2  = (y +3/2)/3  = (z+5/2)/5

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