The equation of the plane passing through the point (1,1,1) and perpendicular to the planes `2x+y-2z=5 and 3x-6y-2z=7`

Options

A. `14x+2y+15x=31`
B. `14x+2y-15z=1`
C. `14x+2y+15x=3`
D. `14x-2y+15z=27`

Answer

A

Explanation

(a) given that required plane is perpendicular to the given two planes ,
therefore , the ormal vector of required plane I sperpendicular to the normal to the given planes .
therefore , the normal vector of required plane is parallel to the vector :
`|{:(hati,hatj,hatk),(2,1,-2),(3,-6,-2):}|=-14hati-2hatj-15hatk`
thus , the equation of required of required plane passing through (1,1,1) will be :
`-14(x-1)-2(y-1)-15(z-1)=0`
`implies 14x+2y+15z=31`