(10 Points) Find the shortest distance from the point (1, 0, -2) to the plane $x + 2y + z = 4$ by minimizing the square of the distance \\
$d(x, y) = \sqrt{(x - 1)^2 + y^2 + (z + 2)^2}$ \\
between (1, 0, -2) and a generic point (x, y, z) on the plane. You will need to recall the quantity \\
$D(a, b) = f_{xx}(a, b)f_{yy}(a, b) - [f_{xy}(a, b)]^2$ \\
in order to determine the points (a, b) that minimize f(x, y).