$\Delta = \frac{1}{{abc}}\,\,\left| {\,\begin{array}{*{20}{c}}{a{b^2}{c^2}}&{abc}&{ab + ac}\\{{a^2}b{c^2}}&{abc}&{bc + ab}\\{{a^2}{b^2}c}&{abc}&{ac + bc}\end{array}\,} \right|$
= $\frac{{{a^2}{b^2}{c^2}}}{{abc}}\,\left| {\,\begin{array}{*{20}{c}}{bc}&1&{ab + ac}\\{ac}&1&{bc + ab}\\{ab}&1&{ac + bc}\end{array}\,} \right|\, = \,abc\,\left| {\,\begin{array}{*{20}{c}}{bc}&1&{\Sigma ab}\\{ac}&1&{\Sigma \,ab}\\{ab}&1&{\Sigma \,ab}\end{array}\,} \right|$
{by ${C_3} \to {C_3} + {C_1}$}
= $abc.\Sigma \,ab\,\left| {\,\begin{array}{*{20}{c}}{bc}&1&1\\{ca}&1&1\\{ab}&1&1\end{array}\,} \right| = 0$, [Since ${C_2} \equiv {C_3}$].
Trick : Put $a = 1,\,b = 2,\,c = 3$ and check it.
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| $\text{X}:$ | $2$ | $3$ | $4$ | $5$ |
| $\text{P}(\text{X}):$ | $\frac{5}{\text{k}}$ | $\frac{7}{\text{k}}$ | $\frac{9}{\text{k}}$ | $\frac{11}{\text{k}}$ |
The value of k is: