Question

Find the equation of the plane passing through the line `(x-1)/5=(y+2)/6=(z-3)/4` and point `(4,3,7)dot`

Answer 1

Any plane through `(x-1)/(5) = (y+2)/(6)=(z-3)/(4)` is
`" "A(x-1)+B(y+2)+C(z-3)=0" "`(i)
where `5A+6B+4C=0" "` (ii)
Also, the plane passes through `(4, 3, 7)`. Therefore,
`" "3A+5B+4C=0" "` (iii)
By (ii) and (iii), `(A)/(4)=(B)/(-8)= (C)/(7)`
Therefore, the plane is
`" "4(x-1)-8(y+2)+7(z-3)=0`
or `" "4x-8y+7z= 41`.