If $$z$$ is a complex number satisfying $$|Re(z)| + |Im(z)| = 4$$, then $$|z|$$ cannot be:

Let $$z = x + iy$$, where $$x = \Re(z)$$ and $$y = \Im(z)$$. The condition given is

$$|x| + |y| = 4.$$

Because the expression contains absolute values, the point $$(x,y)$$ can lie in any of the four quadrants. However, $$|z| = \sqrt{x^{2}+y^{2}}$$ depends only on $$x^{2}$$ and $$y^{2}$$, which are the same in every quadrant. So, without loss of generality, we may restrict ourselves to the first quadrant where $$x \ge 0$$ and $$y \ge 0$$. In that quadrant the equation simplifies to

$$x + y = 4.$$

From this linear relation we can express one variable in terms of the other. Solving for $$y$$ gives

$$y = 4 - x.$$

Now we write the modulus $$|z|$$ in terms of $$x$$ alone. By definition,

$$|z| = \sqrt{x^{2} + y^{2}}.$$

Substituting $$y = 4 - x$$, we obtain

$$|z| = \sqrt{x^{2} + (4 - x)^{2}}.$$

First expand the square:

$$(4 - x)^{2} = 16 - 8x + x^{2}.$$

Adding this to $$x^{2}$$ gives

$$$x^{2} + (4 - x)^{2} = x^{2} + 16 - 8x + x^{2} = 2x^{2} - 8x + 16.$$$

So

$$|z| = \sqrt{2x^{2} - 8x + 16}.$$

To find the range of $$|z|$$ we first find the range of the quadratic expression inside the square root. Denote

$$f(x) = 2x^{2} - 8x + 16,$$

with $$x$$ restricted to $$0 \le x \le 4$$ because $$x + y = 4$$ and both $$x, y \ge 0$$ in the first quadrant.

To locate extrema of $$f(x)$$, we use the calculus fact that a quadratic $$ax^{2} + bx + c$$ has its vertex at $$x = -\tfrac{b}{2a}$$. Here $$a = 2$$ and $$b = -8$$, so

$$x_{\text{vertex}} = -\frac{-8}{2 \cdot 2} = \frac{8}{4} = 2.$$

This $$x = 2$$ lies inside the closed interval $$[0,4]$$, so the quadratic attains its minimum there. Evaluating $$f$$ at the vertex:

$$$f(2) = 2(2)^{2} - 8(2) + 16 = 2 \cdot 4 - 16 + 16 = 8.$$$

Next evaluate $$f(x)$$ at the endpoints.

At $$x = 0$$:

$$$f(0) = 2(0)^{2} - 8(0) + 16 = 16.$$$

At $$x = 4$$:

$$$f(4) = 2(4)^{2} - 8(4) + 16 = 2 \cdot 16 - 32 + 16 = 16.$$$

So the quadratic expression satisfies

$$$8 \le f(x) \le 16 \quad \text{for} \quad 0 \le x \le 4.$$$

Taking square roots, the modulus $$|z|$$ satisfies

$$\sqrt{8} \le |z| \le 4.$$

Therefore any value of $$|z|$$ must lie in the closed interval $$[\sqrt{8},\,4]$$. Equivalently, $$|z|$$ cannot be less than $$\sqrt{8}$$ or greater than $$4$$.

Now examine the four options:

A. $$$\sqrt{\tfrac{17}{2}} = \sqrt{8.5} \approx 2.915 \quad (\text{lies between } \sqrt{8}=2.828\text{ and }4).$$$

B. $$$\sqrt{10} \approx 3.162 \quad (\text{lies between } \sqrt{8}\text{ and }4).$$$

C. $$$\sqrt{7} \approx 2.646 \quad (\text{smaller than } \sqrt{8}).$$$

D. $$$\sqrt{8} = 2.828 \quad (\text{exactly the minimum possible value}).$$$

The only option that falls outside the permissible interval is $$\sqrt{7}$$.

Hence, the correct answer is Option C.