Since, $\left|\begin{array}{lll}x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x\end{array}\right|=0$

Expanding along $R_1$,

$$\begin{array}{rrrl} & x\left(x^2-12\right)-3(2 x-14)+7(12-7 x) & =0 \\ \Rightarrow & x^3-12 x-6 x+42+84-49 x=0 \\ \Rightarrow & x^3-67 x+126=0\quad\text{.... (i)} \end{array}$$

Here, $$126 \times 1=9 \times 2 \times 7$$

For $x=2,2^3-67 \times 2+126=134-134=0$

Hence, $x=2$ is a root.

For $x=7,7^3-67 \times 7+126=469-469=0$

Hence, $x=7$ is also a root.

We have, $\left|\begin{array}{ccc}0 & x y z & x-z \\ y-x & 0 & y-z \\ z-x & z-y & 0\end{array}\right|=\left|\begin{array}{ccc}z-x & x y z & x-z \\ z-x & 0 & y-z \\ z-x & z-y & 0\end{array}\right|\quad$ $$ \left[\because C_1 \rightarrow C_1-C_3\right]$$

$$ =(z-x)\left|\begin{array}{ccc} 1 & x y z & x-z \\ 1 & 0 & y-z \\ 1 & z-y & 0 \end{array}\right|\quad$$ [taking $(z-x)$ common from column 1]

$$\begin{aligned} &\text { Expanding along } R_1 \text {, }\\ &\begin{aligned} & =(z-x)[1 \cdot\{-(y-z)(z-y)\}-x y z(z-y)+(x-z)(z-y)] \\ & =(z-x)(z-y)(-y+z-x y z+x-z) \\ & =(z-x)(z-y)(x-y-x y z) \\ & =(z-x)(y-z)(y-x+x y z) \end{aligned} \end{aligned}$$

Since,

$$\begin{aligned} &f(x)=(1+x)^{17}(1+x)^{23}(1+x)^{41}\left|\begin{array}{ccc} 1 & (1+x)^2 & (1+x)^6 \\ 1 & (1+x)^6 & (1+x)^{11} \\ 1 & (1+x)^2 & (1+x)^6 \end{array}\right|=0\\ &\text { [since, } R_1 \text { and } R_3 \text { are identical] }\\ \therefore\quad &A=0 \end{aligned}$$