= $xyz\left( {1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right)$ $\,\left| {\,\begin{array}{*{20}{c}}1&1&1\\{\frac{1}{y}}&{1 + \frac{1}{y}}&{\frac{1}{y}}\\{\frac{1}{z}}&{\frac{1}{z}}&{1 + \frac{1}{z}}\end{array}\,} \right|$,
by ${R_1} \to {R_1} + {R_2} + {R_3}$
=$xyz\left( {1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right)$$\,\left| {\,\begin{array}{*{20}{c}}1&0&0\\{1/y}&1&0\\{1/z}&0&1\end{array}\,} \right|$, by $\begin{array}{l}{C_2} \to {C_2} - {C_1}\\{C_3} \to {C_3} - {C_1}\end{array}$
= $xyz\left( {1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right)$ $\left| {\,\begin{array}{*{20}{c}}1&0\\0&1\end{array}\,} \right| = xyz\left( {1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z}} \right)$.
Trick: Put $x = 1,\,y = 2$ and $z = 3$, then
$\left| {\,\begin{array}{*{20}{c}}2&1&1\\1&3&1\\1&1&4\end{array}\,} \right| = 2(11) - 1(3) + 1(1 - 3) = 17$
Option $ (a)$ gives, $1 \times 2 \times 3\,\left( {1 + \frac{1}{1} + \frac{1}{2} + \frac{1}{3}} \right) = 17$.
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