If $\quad \overline{\mathrm{a}}=\hat{\mathrm{i}}-\hat{\mathrm{k}}, \overline{\mathrm{b}}=x…
If $\quad \overline{\mathrm{a}}=\hat{\mathrm{i}}-\hat{\mathrm{k}}, \overline{\mathrm{b}}=x \hat{\mathrm{i}}+\hat{\mathrm{j}}+(1-x) \hat{\mathrm{k}} \quad$ and $\overline{\mathrm{c}}=y \hat{\mathrm{i}}+x \hat{\mathrm{j}}+(1+x-y) \hat{\mathrm{k}}$ then $\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})$ depends on
only $x$
only $y$
neither $x$ nor $y$
both $x$ and $y$
Solution
$\left[\begin{array}{lll}
\overline{\mathrm{a}} & \overline{\mathrm{~b}} & \overline{\mathrm{c}}
\end{array}\right]=\left|\begin{array}{ccc}
1 & 0 & -1 \\
x & 1 & 1-x \\
y & x & 1+x-y
\end{array}\right|$ Applying $\mathrm{C}_3 \Rightarrow \mathrm{C}_3+\mathrm{C}_1$, we get
$[\overline{\mathrm{a}} \overline{\mathrm{~b}} \overline{\mathrm{c}}]=\left|\begin{array}{ccc}
1 & 0 & 0 \\
x & 1 & 1 \\
y & x & 1+x
\end{array}\right|=1(1+x-x)=1$