A) only y
B) only\[x\]
C) both\[x\]and y
D) neither\[x\] nor y
Correct Answer: D
Solution :
\[[\overrightarrow{a}\,\overrightarrow{b}\,\overrightarrow{c}]=\left| \begin{matrix} 1 & 0 & -1 \\ x & 1 & 1-x \\ y & x & 1+x-y \\ \end{matrix} \right|\] Applying \[{{C}_{3}}\to {{C}_{3}}+{{C}_{1}}\] \[=\left| \begin{matrix} 1 & 0 & 0 \\ x & 1 & 1 \\ y & x & 1+x \\ \end{matrix} \right|\] \[=1[1+x-x]=1\] Hence,\[[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}]\]neither depends on\[x\]nor y.
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