Let the determinant of a $3 \times 3$ matrix $A$ be $6$. If $B$ is a matrix defined by $B = 5A^2$,then the determinant of $B$ is:

$\left| \begin{array}{ccc} a - b & b - c & c - a \\ x - y & y - z & z - x \\ p - q & q - r & r - p \end{array} \right| = $

Evaluate the determinant: $\left| \begin{array}{ccc} 1/a & 1 & bc \\ 1/b & 1 & ca \\ 1/c & 1 & ab \end{array} \right|$

By using properties of determinants,show that:
$\left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|=(a+b+c)^{3}$

If $\Delta _1 = \left| \begin{matrix} b^5c^6(c^3 - b^3) & a^4c^6(a^3 - c^3) & a^4b^5(b^3 - a^3) \\ b^2c^3(b^6 - c^6) & ac^3(c^6 - a^6) & ab^2(a^6 - b^6) \\ b^2c^3(c^3 - b^3) & ac^3(a^3 - c^3) & ab^2(b^3 - a^3) \end{matrix} \right|$ and $\Delta _2 = \left| \begin{matrix} a & b^2 & c^3 \\ a^4 & b^5 & c^6 \\ a^7 & b^8 & c^9 \end{matrix} \right|$,then $\Delta _1 \Delta _2$ is equal to:

$\left|\begin{array}{ccc}x+y & y+z & z+x \\ z & x & y \\ 1 & 1 & 1\end{array}\right|=$ . . . . . . .