Consider the plane that is parallel to the \( x y \)-plane and goes through the point \( \left(x_{0}, y_{0}, z_{0}\right)=(8,2,5) \). Which position vector is normal to the plane? \( \square[4,0,0] \) Use the normal position vector above and the point \( \left(x_{0}, y_{0}, z_{0}\right) \) to find the equation of the plane. Equation of plane: \( z=0[0,6,0] \)

1. Since the plane is parallel to the \(xy\)-plane, its normal vector must be perpendicular to the \(xy\)-plane. The \(xy\)-plane is horizontal, so the normal points in the vertical direction, i.e. along the \(z\)-axis. Therefore, a normal vector to the plane is \[ \langle 0,0,1 \rangle. \] 2. To find the equation of the plane using the normal vector \(\langle 0,0,1 \rangle\) and the point \((8,2,5)\), we use the point-normal form: \[ a(x-x_0) + b(y-y_0) + c(z-z_0)=0, \] where \(\langle a,b,c \rangle=\langle 0,0,1 \rangle\) and \((x_0,y_0,z_0)=(8,2,5)\). Substituting the values, we get: \[ 0\cdot(x-8) + 0\cdot(y-2) + 1\cdot(z-5)=0. \] This simplifies to: \[ z-5=0, \] and thus the equation of the plane is: \[ z=5. \]