Thank you for reporting, we will resolve it shortly
Q.
If the equations $a(y + z) = x$, $b(z + x)=y$, $c(x+y) = z$ have non-trivial solutions, then the value of $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c} is$
Determinants
Solution:
$\left|\begin{matrix}-1&a&a\\ b&-1&b\\ c&c&-1\end{matrix}\right|=0$
Applying $C_{2} \rightarrow C2 - C_{1}$ and $C_{3} \rightarrow C_{3}-C_{1}$, we get
$\left|\begin{matrix}-1&a+1&a+1\\ b&-\left(b+1\right)&0\\ c&0&-\left(1+c\right)\end{matrix}\right|=0$
Taking $\left(a + 1\right)$, $\left(b + 1\right)$ and $\left(c + 1\right)$ common from $R_{1}$, $R_{2}$ and $R_{3}$ respectively.
$\left|\begin{matrix}-\frac{1}{a+1}&1&1\\ \frac{b}{b+1}&-1&0\\ \frac{c}{c+1}&0&-1\end{matrix}\right|=0$
Expanding along $C_{1}$, we get
$-\frac{1}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}=0$
$\Rightarrow \quad-\frac{1}{a+1}+1-\frac{1}{b+1}+1-\frac{1}{c+1}=0$
$\therefore \quad\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}=2$