Find the equation of the plane perpendicular to the line `(x-1)/2=(y-3)/(-1)=(z-4)/2` and passing through the origin.

Explanation

Any plane passing through the origin is `a(x-0) +b (y-0)+c(z-0)=0`
This is perpendicular to the given line. Therefore, the normal to the plane is parallel to the give line. Thus,
`" "(a)/(2)=(b)/(-1)=(c)/(2)`
Therefore, the required plane is
`" "2(x-0)-1(y-0)+2(z-0)=0`
or `" " 2x-y+2z=0`