Question

If $[(\overline{a}+2\overline{b}+3\overline{c})\times(\overline{b}+2\overline{c}+3\overline{a})]\cdot(\overline{c}+2\overline{a}+3\overline{b})=54$ then the value of $[\overline{a}\overline{b}\overline{c}]$ is

• A. 0
• B. 1
• C. 3
• D. 2

Hint:

Logic Tip: The determinant property is a massive shortcut! Expanding the cross and dot products manually involves multiplying out $3 \times 3 \times 3 = 27$ terms and canceling most of them. Recognizing the coefficient matrix transforms a 5-minute algebraic slog into a 30-second arithmetic calculation.

Answer 1

The Correct Option is C

Concept:
The given expression is a scalar triple product of vectors written as linear combinations of $\vec{a}, \vec{b}, \vec{c}$. \[ [x_1\vec{a} + y_1\vec{b} + z_1\vec{c} \quad x_2\vec{a} + y_2\vec{b} + z_2\vec{c} \quad x_3\vec{a} + y_3\vec{b} + z_3\vec{c}] = \begin{vmatrix} x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{vmatrix} [\vec{a}\ \vec{b}\ \vec{c}] \] 
Step 1: Identify vectors.
\[ \vec{X} = \vec{a} + 2\vec{b} + 3\vec{c} \] \[ \vec{Y} = 3\vec{a} + \vec{b} + 2\vec{c} \] \[ \vec{Z} = 2\vec{a} + 3\vec{b} + \vec{c} \] Given: \[ [\vec{X}\ \vec{Y}\ \vec{Z}] = 54 \] \[ [\vec{X}\ \vec{Y}\ \vec{Z}] = \begin{vmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{vmatrix} [\vec{a}\ \vec{b}\ \vec{c}] \] 
Step 2: Evaluate determinant.
\[ \Delta = \begin{vmatrix} 1 & 2 & 3 \\ 3 & 1 & 2 \\ 2 & 3 & 1 \end{vmatrix} \] \[ = 1(1\cdot1 - 2\cdot3) - 2(3\cdot1 - 2\cdot2) + 3(3\cdot3 - 1\cdot2) \] \[ = 1(1 - 6) - 2(3 - 4) + 3(9 - 2) \] \[ = -5 + 2 + 21 = 18 \] 
Step 3: Find $[\vec{a}\ \vec{b}\ \vec{c}]$.
\[ 18[\vec{a}\ \vec{b}\ \vec{c}] = 54 \] \[ [\vec{a}\ \vec{b}\ \vec{c}] = 3 \] 
Final Answer:
\[ \boxed{3} \]