A man starts walking from the point $$P(-3, 4)$$, touches the x-axis at $$R$$, and then turns to reach at the point $$Q(0, 2)$$. The man is walking at a constant speed. If the man reaches the point Q in the minimum time, then $$50[(PR)^2 + (RQ)^2]$$ is equal to _________.

We have a starting point $$P(-3,4)$$ on the coordinate plane. The man must first reach some point $$R$$ on the x-axis, so we can denote $$R(x,0)$$, and then go to the fixed point $$Q(0,2)$$. The speed of the man is constant, so the time taken is directly proportional to the total distance travelled. Therefore minimising the time is exactly the same as minimising the total distance $$PR+RQ$$.

To minimise a broken-line path that meets a straight line (here, the x-axis) at an intermediate point, we use the well-known reflection idea: reflect one terminal point in the line and join the reflection to the other terminal point by a straight segment. So, we reflect the point $$Q(0,2)$$ in the x-axis. The x-axis is the line $$y=0$$, and reflection in it changes the sign of the y-coordinate while keeping the x-coordinate unchanged. Thus the reflection of $$Q(0,2)$$ is

$$Q'(0,-2).$$

Now the broken path $$P \to R \to Q$$ has exactly the same length as the broken path $$P \to R \to Q'$$ because $$RQ = RQ'$$ (both are vertical mirror images). A straight line is the shortest path between two points, so the shortest possible broken path occurs when $$P,\,R,\,Q'$$ are collinear. Hence $$R$$ must be the point where the straight line through $$P$$ and $$Q'$$ cuts the x-axis.

First we determine the equation of the line through $$P(-3,4)$$ and $$Q'(0,-2)$$. The slope formula is

$$m = \dfrac{y_2 - y_1}{x_2 - x_1}.$$

Substituting $$P(-3,4)$$ as $$(x_1,y_1)$$ and $$Q'(0,-2)$$ as $$(x_2,y_2)$$ we get

$$m = \dfrac{-2 - 4}{\,0 - (-3)\,} = \dfrac{-6}{3} = -2.$$

Using the point-slope form $$y - y_1 = m(x - x_1)$$ with point $$P(-3,4)$$, we have

$$y - 4 = -2\bigl(x - (-3)\bigr) = -2(x + 3).$$

Simplifying,

$$y - 4 = -2x - 6 \quad\Longrightarrow\quad y = -2x - 2.$$

The x-axis is $$y=0$$, so we set $$y=0$$ to find the coordinates of $$R$$:

$$0 = -2x - 2 \quad\Longrightarrow\quad -2x = 2 \quad\Longrightarrow\quad x = -1.$$

Thus

$$R(-1,0).$$

We now calculate the distances $$PR$$ and $$RQ$$ using the distance formula

$$\text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.$$

Distance $$PR$$:

$$PR = \sqrt{(-1 - (-3))^2 + (0 - 4)^2} = \sqrt{(2)^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}.$$

Distance $$RQ$$:

$$RQ = \sqrt{(0 - (-1))^2 + (2 - 0)^2} = \sqrt{(1)^2 + (2)^2} = \sqrt{1 + 4} = \sqrt{5}.$$

Next we compute the quantity requested in the question, namely $$50\bigl[(PR)^2 + (RQ)^2\bigr]$$. First we find the squared lengths:

$$(PR)^2 = (2\sqrt{5})^2 = 4 \times 5 = 20,$$ $$(RQ)^2 = (\sqrt{5})^2 = 5.$$

Adding,

$$(PR)^2 + (RQ)^2 = 20 + 5 = 25.$$

Finally, multiplying by 50:

$$50\bigl[(PR)^2 + (RQ)^2\bigr] = 50 \times 25 = 1250.$$

So, the answer is $$1250$$.