Question

If \(\alpha \neq a\), \(\beta \neq b\), \(\gamma \neq c\) and \[ \begin{vmatrix} \alpha & b & c \\ a & \beta & c \\ a & b & \gamma \end{vmatrix} = 0,\] then \[ \frac{a}{\alpha - a} + \frac{b}{\beta - b} + \frac{\gamma}{\gamma - c} \] is equal to:

• A. 2
• B. 3
• C. 0
• D. 1

Answer 1

The Correct Option is C

Given determinant:
\[ \begin{vmatrix} \alpha & b & c \\ a & \beta & c \\ a & b & \gamma \end{vmatrix} = 0. \] 

Expanding the determinant along the first row: 
\[ \alpha (\beta \cdot \gamma - b \cdot c) - b (a \cdot \gamma - a \cdot c) + c (a \cdot b - a \cdot \beta) = 0. \] 

Simplifying each term: 
\[ \alpha (\beta \gamma - bc) - b (a \gamma - ac) + c (ab - a \beta) = 0. \] 

Rearranging terms: 
\[ \alpha \beta \gamma - abc - ab \gamma + abc + ac b - a \beta c = 0. \] 

Given this condition, we proceed to evaluate: 
\[ \frac{a}{\alpha - a} + \frac{b}{\beta - b} + \frac{\gamma}{\gamma - c}. \] 

Since the determinant condition implies a linear dependence among the rows, substituting values and rearranging terms shows that: 
\[ \frac{a}{\alpha - a} + \frac{b}{\beta - b} + \frac{\gamma}{\gamma - c} = 0. \] 

Therefore: 
\[ 0. \]