1. Key Concepts and Formulas

• Coplanarity of Vectors: Three vectors $\vec{u}$, $\vec{v}$, and $\vec{w}$ are coplanar if their scalar triple product is zero. The scalar triple product can be computed as the determinant of the matrix formed by the components of the vectors. If $\vec{u} = u_1\widehat i + u_2\widehat j + u_3\widehat k$, $\vec{v} = v_1\widehat i + v_2\widehat j + v_3\widehat k$, and $\vec{w} = w_1\widehat i + w_2\widehat j + w_3\widehat k$, then they are coplanar if and only if: $\begin{vmatrix} u_1 & u_2 & u_3 \\ v_1 & v_2 & v_3 \\ w_1 & w_2 & w_3 \end{vmatrix} = 0$
• Geometric Mean (G.M.): For two non-negative numbers $a$ and $b$, their Geometric Mean is given by $\sqrt{ab}$.
• Arithmetic Mean (A.M.): For two numbers $a$ and $b$, their Arithmetic Mean is given by $\frac{a+b}{2}$.
• Harmonic Mean (H.M.): For two non-zero numbers $a$ and $b$, their Harmonic Mean is given by $\frac{2}{\frac{1}{a} + \frac{1}{b}} = \frac{2ab}{a+b}$.

2. Step-by-Step Solution

Step 1: Identify the given vectors and the condition for coplanarity. We are given three vectors: $\vec{v_1} = a\widehat i + a\widehat j + c\widehat k$ $\vec{v_2} = 1\widehat i + 0\widehat j + 1\widehat k$ $\vec{v_3} = c\widehat i + c\widehat j + b\widehat k$ The problem states that these vectors lie in a plane, which means they are coplanar. The condition for coplanarity is that their scalar triple product is zero.

Step 2: Form the determinant for the scalar triple product. The scalar triple product of $\vec{v_1}$, $\vec{v_2}$, and $\vec{v_3}$ is given by the determinant of the matrix formed by their components: $\begin{vmatrix} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{vmatrix}$ Since the vectors are coplanar, this determinant must be equal to zero. $\begin{vmatrix} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{vmatrix} = 0$

Step 3: Simplify the determinant using row/column operations. To simplify the calculation, we can use column operations. Let's perform the operation $C_1 \to C_1 - C_2$ (replace the first column with the first column minus the second column): $\begin{vmatrix} a-a & a & c \\ 1-0 & 0 & 1 \\ c-c & c & b \end{vmatrix} = \begin{vmatrix} 0 & a & c \\ 1 & 0 & 1 \\ 0 & c & b \end{vmatrix}$ The determinant is now equal to zero: $\begin{vmatrix} 0 & a & c \\ 1 & 0 & 1 \\ 0 & c & b \end{vmatrix} = 0$

Step 4: Expand the simplified determinant. We can expand this determinant along the first column, as it contains two zeros, which will simplify the calculation: $0 \cdot \begin{vmatrix} 0 & 1 \\ c & b \end{vmatrix} - 1 \cdot \begin{vmatrix} a & c \\ c & b \end{vmatrix} + 0 \cdot \begin{vmatrix} a & c \\ 0 & 1 \end{vmatrix} = 0$ This simplifies to: $-1 \cdot (a \cdot b - c \cdot c) = 0$ $-(ab - c^2) = 0$ $c^2 - ab = 0$

Step 5: Solve the equation for c. Rearranging the equation, we get: $c^2 = ab$ Taking the square root of both sides, we have: $c = \pm\sqrt{ab}$ The problem states that $a$, $b$, and $c$ are distinct non-negative numbers. Since $c$ must be non-negative, we take the positive square root: $c = \sqrt{ab}$

Step 6: Relate the result to the given options. The expression $c = \sqrt{ab}$ is the definition of the Geometric Mean of the two non-negative numbers $a$ and $b$.

3. Common Mistakes & Tips

• Mistake 1: Sign Errors in Determinant Expansion: When expanding determinants, especially $3 \times 3$ matrices, be very careful with the alternating signs ($+,-,+$ for expansion along a row or column). Using column/row operations to introduce zeros can significantly reduce the chance of such errors.
• Mistake 2: Forgetting Non-Negativity: The problem specifies that $a, b, c$ are non-negative. This is crucial when taking the square root of $c^2 = ab$, as it restricts $c$ to be non-negative.
• Tip 1: Strategic Simplification: Always look for opportunities to simplify determinants using row or column operations before expanding. Operations like making elements zero or making elements identical can make the expansion much easier.
• Tip 2: Definition of Means: Keep the definitions of Arithmetic Mean, Geometric Mean, and Harmonic Mean in mind. These often appear in problems involving algebraic relationships.

4. Summary

The problem requires us to find the relationship of $c$ with $a$ and $b$ given that three vectors are coplanar. The condition of coplanarity for three vectors is that their scalar triple product is zero. By setting up the scalar triple product as a determinant and simplifying it, we arrived at the equation $c^2 = ab$. Since $c$ is non-negative, this implies $c = \sqrt{ab}$, which is the definition of the Geometric Mean of $a$ and $b$.

The final answer is $\boxed{A}$ which corresponds to the Geometric Mean of $a$ and $b$.