For $$z \in \mathbb{C}$$, if the minimum value of $$(|z - 3\sqrt{2}| + |z - p\sqrt{2}i|)$$ is $$5\sqrt{2}$$, then a value of $$p$$ is

We need the value of $$p$$ such that the minimum of $$|z - 3\sqrt{2}| + |z - p\sqrt{2}\,i|$$ is $$5\sqrt{2}$$.

Geometric interpretation.

The expression $$|z - 3\sqrt{2}| + |z - p\sqrt{2}\,i|$$ is the sum of distances from $$z$$ to two fixed points in the complex plane:

$$A = 3\sqrt{2}$$ (on the real axis) and $$B = p\sqrt{2}\,i$$ (on the imaginary axis).

Apply the triangle inequality.

By the triangle inequality, the minimum value of $$|z - A| + |z - B|$$ equals the distance $$|A - B|$$, achieved when $$z$$ lies on the line segment $$AB$$.

Compute $$|A - B|$$.

$$|A - B| = |3\sqrt{2} - p\sqrt{2}\,i| = \sqrt{(3\sqrt{2})^2 + (p\sqrt{2})^2} = \sqrt{18 + 2p^2}$$

Solve for $$p$$.

$$\sqrt{18 + 2p^2} = 5\sqrt{2}$$

Squaring both sides:

$$18 + 2p^2 = 50 \implies 2p^2 = 32 \implies p^2 = 16 \implies p = \pm 4$$

Select from the options.

Among the given options, $$p = 4$$ is available.

The answer is Option D: $$4$$.