A) \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-3abc\]
B) \[3ab\]
C) \[3a+5b\]
D) 0
Correct Answer: D
Solution :
\[\left| \,\begin{matrix} a+b & a+2b & a+3b \\ a+2b & a+3b & a+4b \\ a+4b & a+5b & a+6b \\ \end{matrix}\, \right|\,=\,\left| \,\begin{matrix} a+b & a+2b & a+3b \\ b & b & b \\ 2b & 2b & 2b \\ \end{matrix}\, \right|\] = 0 \[\left\{ \text{by }\begin{matrix} {{R}_{2}}\to {{R}_{2}}-{{R}_{1}} \\ {{R}_{3}}\to {{R}_{3}}-{{R}_{2}} \\ \end{matrix} \right\}\] Trick: Putting\[a=1=b\]. The determinant will be\[\left| \,\begin{matrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 5 & 6 & 7 \\ \end{matrix}\, \right|=0\]. Obviously answer is Note: Students remember while taking the values of \[a,\,b,\,\,c,.......\] that for their values, the options , , and should not be identical.
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