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If \(\alpha \neq a , \beta \neq b , \gamma \neq c\) and \(\left|\begin{array}{lll}\alpha & b & c \\ a & \bet…

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If \(\alpha \neq a , \beta \neq b , \gamma \neq c\) and \(\left|\begin{array}{lll}\alpha & b & c \\ a & \beta & c \\ a & b & \gamma\end{array}\right|=0\), then \(\frac{ a }{\alpha- a }+\frac{ b }{\beta- b }+\frac{\gamma}{\gamma- c }\) is equal to:

[JEE Main 2024, 08 Apr (Shift 2)]

a

\(2\)

b

\(3\)

c

\(0\)

d

\(1\)

✓ Correct answer: c)

\(0\)

Explanation

Solution of determinant by using operation of determinant

\({R}_{1}\to {R}_{1}-{R}_{2}\\ {R}_{2}\to {R}_{2}-{R}_{3}\)

\(\left|\begin{matrix}\alpha -a & b-\beta & 0 \\ 0 & \beta -b & c-\gamma \\ a & b & \gamma \end{matrix}\right|=0\)

\(\left(\alpha -a\right)\left(\beta -b\right)\left(\gamma -c\right)\left|\begin{matrix}1 & -1 & 0 \\ 0 & 1 & -1 \\ \frac{a}{\alpha -a} & \frac{b}{\beta -b} & \frac{\gamma }{\gamma -c}\end{matrix}\right|=0\)

\(\frac{a}{\alpha -a}\left(1-0\right)-\frac{b}{\beta -b}\left(-1-0\right)+\frac{\gamma }{\gamma -c}\left(1-0\right)=0\)

\(\frac{a}{\alpha -a}+\frac{b}{\beta -b}+\frac{\gamma }{\gamma -c}=0\)

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