Find the equation of the plane passing through the point `( 1, 3, 2)`and perpendicular to each of the planes `x + 2y + 3z = 5`and `3x + 3y + z = 0`.

Explanation

Given planes are
`" " x+2y+3z=5" "`(i)
`" "3x+3y+z=0" "`(ii)
Normal to planes (i) and (ii) are `hati +2hatj+3hatk and 3hati+3hatj+hatk`, respectively.
Required plane is perpendicular to planes (i) and (ii), i.e., to perpendicular to the line of intersection of planes.
Vector along the line of intersection of planes (i) and (ii) is `|{:(hati,,hatj,,hatk),(1,,2,,3),(3,,3,,1):}|=-7hati+8hatj-3hatk`
Therefore, normal to given plane is `-7hati+8hatj-3hatk`.
Plane is passing through the point `(-1, 3, 2)`
The equation of required plane is
`" "-7(x+1)+8(y-3)-3(z-2)=0`
or `" "7x-3y+8z+25=0`.
Also the plane is passing through the point `(1, 2, 0)` , hence, the equation of the plane is
`" "0(x-1)+3(y-2)+3(z-0)=0`
or `" "y+z-2=0`.