If a complex number $$z$$ satisfies the equation $$x + \sqrt{2}|z + 1| + i = 0$$, then $$|z|$$ is equal to :

Let the complex number be $$z = x + iy$$, where $$x$$ and $$y$$ are real.

The given equation is $$z + \sqrt{2}\,|\,z + 1\,| + i = 0$$

Substituting $$z = x + iy$$ gives $$(x + iy) + \sqrt{2}\,|\,z + 1\,| + i = 0$$ Combine the real and imaginary parts:

Real part : $$x + \sqrt{2}\,|\,z + 1\,| = 0$$ $$-(1)$$
Imaginary part : $$y + 1 = 0 \;\Longrightarrow\; y = -1$$ $$-(2)$$

Next, write $$|\,z + 1\,|$$ in terms of $$x$$ and $$y$$. Since $$z + 1 = (x + 1) + iy$$, using $$y = -1$$ from $$(2)$$ gives

$$|\,z + 1\,| = \sqrt{(x + 1)^2 + (-1)^2} = \sqrt{(x + 1)^2 + 1}$$ $$-(3)$$

Insert $$(3)$$ into the real-part equation $$(1)$$:

$$x + \sqrt{2}\,\sqrt{(x + 1)^2 + 1} = 0$$ $$\Longrightarrow$$ $$x = -\sqrt{2}\,\sqrt{(x + 1)^2 + 1}$$

Square both sides to remove the square root:

$$x^2 = 2\bigl[(x + 1)^2 + 1\bigr]$$

Simplify the right side: $$(x + 1)^2 + 1 = x^2 + 2x + 1 + 1 = x^2 + 2x + 2$$ Hence

$$x^2 = 2(x^2 + 2x + 2) = 2x^2 + 4x + 4$$

Bring all terms to one side:

$$0 = 2x^2 + 4x + 4 - x^2 = x^2 + 4x + 4$$

Recognise the perfect square: $$(x + 2)^2 = 0 \;\Longrightarrow\; x = -2$$

Thus, from $$(2)$$ and the value of $$x$$, $$z = -2 - i$$

The magnitude of $$z$$ is

$$|z| = \sqrt{(-2)^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5}$$

Therefore, $$|z| = \sqrt{5}$$, which corresponds to Option C.