Optimization Problem

The distance from a point ${\bf (x,y,z) }$ in the plane ${\bf x + 2y - z = 3 }$ to the origin is ${\bf d = \sqrt{x^2 + y^2 + z^2} }$ , and since d cannot be negative, it's a minimum when ${\bf w = d^2 = x^2 + y^2 = z^2}$ . $$

Using Lagrangian multipliers ${\bf \implies F(x,y,z) = x^2 + y^2 + z^2 + \lambda(x + 2y - z - 3) \implies}$

${\bf \frac{\partial F}{\partial x} = 2x + \lambda = 0 }$ $$

${\bf \frac{\partial F}{\partial y} = 2y + 2\lambda = 0 }$ $$

${\bf \frac{\partial F}{\partial z} = 2z - \lambda = 0 }$ $$

${\bf (*) \frac{\partial F}{\partial \lambda} = x + 2y - z - 3 = 0}$

$$ ${\bf \implies \lambda = -2x = -y = 2z \implies y = 2x \: and \: z = -x }$

Replacing this in ${\bf (*) \implies x = \frac{1}{2} \implies y = 1 \: and \: z = \frac{-1}{2} }$

${\bf \therefore }$ the point on the given plane closet to the origin is:

${\bf (\frac{1}{2},1,\frac{-1}{2}) = (a,b,c) \implies a + b + c = 1. }$