If $A$ is a matrix of order $3 \times 3$, then $\left(A^2\right)^{-1}=\left(A^{-1}\right)^2$.

If $A$ is a matrix of order $3 \times 3$, then the number of minors in determinant of $A$ are 9 . [since, in a $3 \times 3$ matrix, there are 9 elements]

The sum of products of elements of any row with the cofactors of corresponding elements is equal to value of the determinant.

$$\text { Let } \Delta=\left|\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right|$$

Expanding along $R_1$,

$$\begin{aligned} & \Delta=a_{11} A_{11}+a_{12} A_{12}+a_{13} A_{13} \\ &=\text { Sum of products of elements of } R_1 \text { with their } \\ & \text { corresponding cofactors } \end{aligned}$$

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]

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