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: \( z=0[0,6,0] \)

The normal vector for a plane parallel to the \( xy \)-plane will be directed along the \( z \)-axis, which can be represented as \( \mathbf{n} = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \). This means that any plane parallel to the \( xy \)-plane will have the form \( z = k \). Since the plane in question itself passes through the point \( (8, 2, 5) \), we can simply conclude that the equation of the plane is \( z = 5 \). This means every point on this plane has the same \( z \)-coordinate of 5, allowing for any \( x \) and \( y \) values.