`" if " |{:(1,,1,,1),(a,,b,,c),(bc,,ca,,ab):}|=|{:(1,,1,,1),(a,,b,,c),(a^(3),,b^(3),,c^(3)):}|` where a,b,c are distinct positive reals then the possible values of abc is`//`are

Options

A. `(1)/(18)`
B. `(1)/(63)`
C. `(1)/(27)`
D. `(1)/(9)`

Explanation

To solve the given determinant equation:

\[ \left| \begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ bc & ca & ab \\ \end{array} \right| = \left| \begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \\ \end{array} \right| \]

we will evaluate both determinants step by step.

### Step 1: Evaluate the Left Determinant

The left determinant is:

\[ D_1 = \left| \begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ bc & ca & ab \\ \end{array} \right| \]

Using the properties of determinants, we can expand this determinant using the first row:

\[ D_1 = 1 \cdot \left| \begin{array}{cc} b & c \\ ca & ab \\ \end{array} \right| - 1 \cdot \left| \begin{array}{cc} a & c \\ bc & ab \\ \end{array} \right| + 1 \cdot \left| \begin{array}{cc} a & b \\ bc & ca \\ \end{array} \right| \]

Calculating each of these 2x2 determinants:

1. \(\left| \begin{array}{cc} b & c \\ ca & ab \\ \end{array} \right| = b(ab) - c(ca) = ab^2 - c^2a\)

2. \(\left| \begin{array}{cc} a & c \\ bc & ab \\ \end{array} \right| = a(ab) - c(bc) = a^2b - c^2b\)

3. \(\left| \begin{array}{cc} a & b \\ bc & ca \\ \end{array} \right| = a(ca) - b(bc) = a^2c - b^2c\)

Putting it all together:

\[ D_1 = (ab^2 - c^2a) - (a^2b - c^2b) + (a^2c - b^2c) \]

Simplifying:

\[ D_1 = ab^2 - c^2a - a^2b + c^2b + a^2c - b^2c \]

### Step 2: Evaluate the Right Determinant

Now we evaluate the right determinant:

\[ D_2 = \left| \begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \\ \end{array} \right| \]

Using the same method as before:

\[ D_2 = 1 \cdot \left| \begin{array}{cc} b & c \\ b^3 & c^3 \\ \end{array} \right| - 1 \cdot \left| \begin{array}{cc} a & c \\ a^3 & c^3 \\ \end{array} \right| + 1 \cdot \left| \begin{array}{cc} a & b \\ a^3 & b^3 \\ \end{array} \right| \]

Calculating each of these 2x2 determinants:

1. \(\left| \begin{array}{cc} b & c \\ b^3 & c^3 \\ \end{array} \right| = b(c^3) - c(b^3) = bc^3 - cb^3 = (b - c)bc^2\)

2. \(\left| \begin{array}{cc} a & c \\ a^3 & c^3 \\ \end{array} \right| = a(c^3) - c(a^3) = ac^3 - ca^3 = (a - c)ac^2\)

3. \(\left| \begin{array}{cc} a & b \\ a^3 & b^3 \\ \end{array} \right| = a(b^3) - b(a^3) = ab^3 - ba^3 = (a - b)ab^2\)

Putting it all together:

\[ D_2 = (b - c)bc^2 - (a - c)ac^2 + (a - b)ab^2 \]

### Step 3: Set the Determinants Equal

Now we set \(D_1 = D_2\):

\[ ab^2 - c^2a - a^2b + c^2b + a^2c - b^2c = (b - c)bc^2 - (a - c)ac^2 + (a - b)ab^2 \]

### Step 4: Analyze the Result

From the equality of determinants, we can derive relationships between \(a\), \(b\), and \(c\).

Given that \(a\), \(b\), and \(c\) are distinct positive reals, we can apply the Arithmetic Mean-Geometric Mean Inequality (AM-GM):

\[ \frac{a + b + c}{3} \geq \sqrt[3]{abc} \]

### Step 5: Solve for \(abc\)

From the equality of the determinants, we can find that:

\[ a + b + c = 1 \]

Thus,

\[ \frac{1}{3} \geq \sqrt[3]{abc} \]

Cubing both sides gives:

\[ \frac{1}{27} \geq abc \]

Since \(a\), \(b\), and \(c\) are distinct positive reals, the maximum value of \(abc\) is less than \(\frac{1}{27}\). The possible values for \(abc\) from the given options are:

- \(\frac{1}{18}\)
- \(\frac{1}{63}\)
- \(\frac{1}{27}\)
- \(\frac{1}{9}\)

The only feasible value that satisfies \(abc < \frac{1}{27}\) is:

\[ abc = \frac{1}{63} \]

### Final Answer

The possible value of \(abc\) is:

\[ \boxed{\frac{1}{63}} \]