Find the equation of a plane which passes through the point (3, 2, 0) and contains the line `(x-3)/1=(y-6)/5=(z-4)/4`

Options

A. `x-y+z=1`
B. `x+y+z=5`
C. `x+2y-z=1`
D. `2x-y+z=5`

Answer

A

Explanation

The equation of a plane containing the line
`(x-3)/1=(y-6)/5=(z-4)/4` is ltbr. `a(x-3)+b(y-6)+c(z-4)=0`…………..i
where `a+5b+4c=0`…………..ii
This plane will pass through (3,2,0) if
`a(0)+ b(-4)+c(-4)=0`…………….iii
Solving i and iii we get
`a/(-20+16)=b/(0+4)=c/(-4+0)=a/1=b/(-1)=c/1`
Putting the values of a,b,c in i we obtain that the equation of the plane is `x-y+z-1=0`.