The equation of the plane passing through the line of intersection of the planes `x+y+z=1,2x+3y+4z=7`, and perpendicular to the plane `x-5y+3z=5` is given by

Options

A. `x+2y+3z-6=0`
B. `x+2y+3z+6=0`
C. `3x+4y+5z-8=0`
D. `3x+4y+5z+8=0`

Answer

A

Explanation

Given planes, `p_1:x+y+z=1`
`p_2: 2x+3y+4z=7`
So, equation of plane passing through intersection of planes `p_1 and p_2` is
`x+y+z-1+k(2x+3y+4z-7)=0`
`rArr x+y+z-1+2kx+3ky+4kz-7k=0`
`rArr x(1+2k)+y(1+3k)+z(1+4k)-1-7k=0`
This is perpendicular to `x-5y+3z=5`
`rArr x-5y+3z-5=0`
`rArr 1(1+2k)-5(1+3k)+(1+4k)=0`
`rArr 1+2k-5-15k+3+12k=0`
`rArr -k-1=0 rArr k=-1`
`:."Equation of plane is "x+y+z-1-1(2x+3y+4z-7)=0`
`rArr x+y+z-1-2x-3y-4z+7=0`
`-x-2y-3z+6=0`
`rArr x+2y+3z-6=0`.