Let $$a, b$$ and $$c$$ be distinct non-negative numbers. If the vectors $$a\hat{i} + a\hat{j} + c\hat{k}, \hat{i} + \hat{k}$$ and $$c\hat{i} + c\hat{j} + b\hat{k}$$ lie in a plane, then $$c$$ is

Let the three given vectors be
$$\vec{v}_1 = a\hat{i}+a\hat{j}+c\hat{k} = (a,\,a,\,c),$$
$$\vec{v}_2 = \hat{i}+\hat{k} = (1,\,0,\,1),$$
$$\vec{v}_3 = c\hat{i}+c\hat{j}+b\hat{k} = (c,\,c,\,b).$$

For $$\vec{v}_1,\,\vec{v}_2,\,\vec{v}_3$$ to lie in the same plane, their scalar triple product must vanish:

$$[\vec{v}_1\,\vec{v}_2\,\vec{v}_3] \;=\; 0.$$

Write the scalar triple product as a determinant:

$$\begin{vmatrix} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{vmatrix} = 0.$$

Evaluate the determinant:

$$$ \begin{aligned} \begin{vmatrix} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{vmatrix} &= a\begin{vmatrix}0 & 1 \\ c & b\end{vmatrix} \;-\; a\begin{vmatrix}1 & 1 \\ c & b\end{vmatrix} \;+\; c\begin{vmatrix}1 & 0 \\ c & c\end{vmatrix} \\[4pt] &= a(0\cdot b - 1\cdot c) \;-\; a(1\cdot b - 1\cdot c) \;+\; c(1\cdot c - 0\cdot c) \\[4pt] &= a(-c) - a(b - c) + c^2 \\[4pt] &= -ac - a b + ac + c^2 \\[4pt] &= c^2 - a b. \end{aligned} $$$

The coplanarity condition therefore gives

$$c^2 - a b = 0 \;\;\Longrightarrow\;\; c^2 = a b.$$\

Since $$a, b, c$$ are non-negative and distinct, we take the positive square root:

$$c = \sqrt{ab},$$

which is exactly the geometric mean of $$a$$ and $$b$$.

Hence the correct option is:
Option A which is: the Geometric Mean of $$a$$ and $$b$$.