Question
Without expanding the determinants, show that
$\left|\begin{array}{ccc}x a & y b & z c \\ a^2 & b^2 & c^2 \\ 1 & 1 & 1\end{array}\right|$$=\left|\begin{array}{ccc}x & y & z \\ a & b & c \\ b c & c a & a b\end{array}\right|$
$\left|\begin{array}{ccc}x a & y b & z c \\ a^2 & b^2 & c^2 \\ 1 & 1 & 1\end{array}\right|$$=\left|\begin{array}{ccc}x & y & z \\ a & b & c \\ b c & c a & a b\end{array}\right|$
Taking a, b, c common from $\mathrm{C}_1, \mathrm{C}_2, \mathrm{C}_3$
respectively, we get
L.H.S. $=a b c\left|\begin{array}{ccc}x & y & z \\ a & b & c \\ \frac{1}{a} & \frac{1}{b} & \frac{1}{c}\end{array}\right|$
$=\left|\begin{array}{ccc}x & y & z \\ a & b & c \\ \frac{a b c}{a} & \frac{a b c}{b} & \frac{a b c}{c}\end{array}\right|$
$=\left|\begin{array}{ccc}x & y & z \\ a & b & \mathrm{c} \\ \mathrm{bc} & c a & a b\end{array}\right|=$ R.H.S.
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of getting both the balls green, when two balls are selected at random is $\frac{1}{7}$, find $m$.