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)/4dot`

Options

A. `x -y + z =1`
B. `x + y + z = 5`
C. `x + 2y -x = 0`
D. `2x - y + z = 5`

Answer

A

Explanation

Any plane passing through the line (3,2,0) is
`a(x-3)+b(y-2)+c(z-0) = 0 " "……(i)`
Plane is passing throught the line `(x -3)/(1) =(y-6)/(5) = (z-4)/(4)`
`rArr a(3-3) + b(6-2)+c(4-0) = 0`
`rArr " "0a+4b + 4c = 0" "......(ii)`
Since, the given plane is passing through the line, therefore the DR's of the normal is perpendicular to the line.
`therefore " "a + 5b + 4c = 0 " ".......(iii)`
On solving Eqs. (ii) and(iii) , we get `(a)/(16-20) = (b)/(4-0) = (c)/(0-4)`
`rArr (a)/(-1) = (b)/(1) =(c)/(-1)`
On putting the values of a, b and c in Eqs. (i), we get `x -y + z = 1`