Let $$ABC$$ be a triangle in the $$xy$$ plane, where $$B$$ is at the origin $$(0,0)$$. Let $$BC$$ be produced to $$D$$ such that $$BC:CD=1:1$$, $$CA$$ be produced to $$E$$ such that $$CA:AE=1:2$$ and $$AB$$ be produced to $$F$$ such that $$AB:BF=1:3$$. Let $$G(32,24)$$ be the centroid of the triangle $$ABC$$ and $$K$$ be the centroid of the triangle $$DEF$$. Find the length $$GK$$.

Let the coordinates of the vertices be
$$A(x_A,\,y_A),\qquad B(0,0),\qquad C(x_C,\,y_C).$$

Step 1: Coordinates of the auxiliary points

1. $$BC:CD = 1:1$$ and $$D$$ lies beyond $$C$$ on $$BC$$.
 Vector form $$\vec{D} = \vec{C} + (\vec{C}-\vec{B}) = 2\vec{C}$$
 Hence $$D(2x_C,\,2y_C).$$

2. $$CA:AE = 1:2$$ with $$E$$ beyond $$A$$ on $$CA$$.
 Vector form $$\vec{E} = \vec{A} + 2(\vec{A}-\vec{C}) = 3\vec{A}-2\vec{C}$$
 Hence $$E(3x_A-2x_C,\,3y_A-2y_C).$$

3. $$AB:BF = 1:3$$ with $$F$$ beyond $$B$$ on $$AB$$.
 Vector form $$\vec{F} = \vec{B} + 3(\vec{B}-\vec{A}) = -3\vec{A}$$
 Hence $$F(-3x_A,\,-3y_A).$$

Step 2: Use the centroid of $$\triangle ABC$$

The centroid formula gives
$$\vec{G}=\frac{\vec{A}+\vec{B}+\vec{C}}{3} = \frac{\vec{A}+\vec{C}}{3}.$$
Given $$G(32,24),$$ so

$$\vec{A}+\vec{C}=3\vec{G}=(96,\,72). \quad -(1)$$

Step 3: Centroid of $$\triangle DEF$$

Add the position vectors of $$D,E,F$$:

$$\vec{D}+\vec{E}+\vec{F} = 2\vec{C} + (3\vec{A}-2\vec{C}) + (-3\vec{A}) = 0.$$

Therefore the centroid of $$\triangle DEF$$ is

$$\vec{K}= \frac{\vec{D}+\vec{E}+\vec{F}}{3} = \vec{0},$$
so $$K(0,0),$$ the origin.

Step 4: Distance between $$G(32,24)$$ and $$K(0,0)$$

$$GK = \sqrt{(32-0)^2 + (24-0)^2} = \sqrt{32^2 + 24^2} = \sqrt{1024 + 576} = \sqrt{1600} = 40.$$

Hence the required length is 40.

Final Answer: 40