The vector equation of the plane through the line of intersection of the planes x+y+z=1 and 2 x+3 y+4 z=5 , which is perpendicular to the plane x-y+z=0 , is

Step-by-step solution

The equation of the required plane aligned & (x+y+z-1)+ (2 x+3 y+4 z-5)=0 & (1+2 ) x+(1+3 ) y+(1+4 ) z-1-5 =0 aligned Let a , b , c be the d.r.s. of the required plane. From (i), a=1+2 , b=1+3 , c =1+4 The required plane is perpendicular to x-y+z=0 aligned & a-b+c=0 & 1+2 -(1+3 )+1+4 =0 & 1+3 =0 & =- 1 3 aligned Substituting =- 1 3 in (i), we get aligned & (1- 2 3 ) x+ (1- 3 3 ) y+ (1- 4 3 ) z-1+ 5 3 =0 & x-z+2=0 aligned Its vector equation is r ( i - k )+2=0