Point closest to the origin on the line of two intersecting planes

Question

So I understand that the distance from the origin $(0,0,0)$ to a point $(x,y,z)$ is:

$$ \sqrt {x^2 + y^2 + z^2} $$

How would I go about finding the point closest to the origin $(0,0,0)$ on the line of intersection of the planes $$2z + y = 12$$ $$x + y = 6$$

Using the method of Lagrange multipliers. Then, calculating the distance between the origin and the point? How would I go about doing this?

I have parameterized like so, but don't know how to continue:

\begin{eqnarray} z &=& 3+\frac{t}{2}\\ x &=& t\\ y &=& 6-t \end{eqnarray}

Answer 1

You can instead minimize the squared distance to simplify taking derivatives since squaring is an increasing transformation (minimizing the distance is the same as minimizing the distance squared).

$$\mathcal{L} = x^2 +y^2 + z^2 + \lambda (2z+y-12) + \mu (x+y-6)$$

Then find $x,y,z$ minimizing this in terms of the lagrange multipliers. Then substitute those values back into the lagrangian above, and finally optimize the lagrange multipliers.

Answer 2

Then you are after the minimum of $t^2+(6-t)^2+\left(\frac t2+3\right)^2=\frac94t^2-9t+45$. Can you take it from here?

Answer 3

$$\sqrt{x^2+y^2+z^2}=\sqrt{(-6+2z)^2+(12-2z)^2+z^2}=$$ $$=3\sqrt{z^2-8z+20}=3\sqrt{(z-4)^2+4}\geq6.$$ The equality occurs for $z=4,$ which gives the needed point: $(2,4,4).$