Find the equation of the plane containing the lines `(x-5)/4=(y-7)/4=(z+3)/(-5)a n d(x-8)/7=(y-4)/1=(z-5)/3dot`

Options

A. `17x-47y-24z+152=0`
B. `17x+47y-24z+172=0`
C. `17x+47y+24z+172=0`
D. `17x-47y+24z+172=0`

Answer

A

Explanation

The equation of plane, in which the
`"line"(x-5)/(4)=(y-7)/(4)=(z+3)/(-5)` lies is
`A(x-5)+B(y-7)+C(z+3)=0" "...(i)`
where A, B and C are the direction ratios of the plane. Since, the first line lie on the plane.
`:.` Direction ratios of normal to the plane is perpendicular to the direction ratios of line i.e.
`4A+4B-5C=0" "...(ii)`
Since, line `(x-8)/(7)=(y-4)/(1)=(z-5)/(3)` lies in this plane. The direction ratios is also perpendicular to this line.
`:." "7A9+B+3C=0" "...(iii)`
From Eqs. (ii) and (iii), we get
`(A)/(17)=(B)/(-47)=(C)/(-24)`
`:.17(x-5)-47(y-7)-24(z+3)=0`
`implies" "17x-47y-24z+152=0`