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. (0,3,1)
B. (0,7,-10)
C. (0,-2,1)
D. (0,7,-10)

Answer

D

Explanation

Equation of plane through the point (3,2,0) and the line `(x-1)/(1)=(y-2)/(5)=(z-3)/(4)` lies on it, is
`|{:(x-x_(1),y-y_(1),z-z_(1)),(x_(2)-x_(1),y_(2)-y_(1),z_(2)-z_(1)),(a,b,c):}|=0`
`rArr" "|{:(x-1,y-2,z-3),(2,0,-3),(1,5,4):}|=0`
`rArr(x-1)(15)-(y-2)(8+3)+(z-3)(10)=0`
`rArr" "15x-15-11y+22+10z-30=0`
`rArr" "15x-11y+10z=23`
is required equation of plane.
The point (0,7,10) satisfies above equation of plane hence it contains the plane.