Q55 / 90 · JEE Main 2012 (12 May Online) · Mathematics
If $\left|\begin{array}{ccc}-2 a & a+b & a+c \\ b+a & -2 b & b+c \\ c+a & b+c & -2 c\end{array}\right|$
$$
=\alpha(a+b() b+c() c+a) \neq 0
$$
then $\alpha$ is equal to
- A$a+b+c$
- B$a b c$
- Correct answerC4
- D1
Solution
Let $\Delta=\left|\begin{array}{ccc}-2 a & a+b & a+c \\ b+a & -2 b & b+c \\ c+a & b+c & -2 c\end{array}\right|$
Applying $C_1+C_3$ and $C_2+C_3$
$$
\Delta=\left|\begin{array}{ccc}
-a+c & 2 a+b+c & a+c \\
2 b+a+c & -b+c & b+c \\
a-c & b-c & -2 c
\end{array}\right|
$$
Now, applying $R_1+R_3$ and $R_2+R_3$
$$
\Delta=\left|\begin{array}{ccc}
0 & 2(a+b & ) a-c \\
2(a+b & 0 & b-c \\
a-c & b-c & -2 c
\end{array}\right|
$$
On expanding, we get
$$
\begin{aligned}
\Delta= & -2(a+b)\{-2 c[2(a+b)] \\
& -(a-c)(b-c)\} \\
& +(a-c)[2(a+b)(b-c)] \\
\Delta= & 8 \mathrm{c}(a+b)(a+b) \\
& +4(a+b)(a-c)(b-c)
\end{aligned}
$$
$$
\begin{aligned}
& =4(a+b)\left[2 a c+2 b c+a b-b c-a c+c^2\right] \\
& =4(a+b)\left[a c+b c+a b+c^2\right] \\
& =4(a+b)[c(a+c)+b(a+c)] \\
& =4(a+b)(b+c)(c+a) \\
& =\alpha(a+b)(b+c)(c+a)
\end{aligned}
$$
Hence, $\alpha=4$
Hence, the correct answer is option: 'C'.