If $\mathbf{a}+x \mathbf{b}+y \mathbf{c}=\mathbf{0}$, $\mathbf{a} \times \mathbf{b}+\mathbf{b} \times…
If $\mathbf{a}+x \mathbf{b}+y \mathbf{c}=\mathbf{0}$, $\mathbf{a} \times \mathbf{b}+\mathbf{b} \times \mathbf{c}+\mathbf{c} \times \mathbf{a}=6(\mathbf{b} \times \mathbf{c})$, then the locus of the point $(x, y)$ is
$x^2+y^2=1$
$x+y-5=0$
$2 x+6 y=5$
$x+y+6=0$
Solution
Given,
$
\mathbf{a}+x \mathbf{b}+y \mathbf{c}=\mathbf{0}
$
As we know that, if $\mathbf{a}+\mathbf{b}+\mathbf{c}=0$
Then, $\mathbf{a} \times \mathbf{b}=\mathbf{b} \times \mathbf{c}=\mathbf{c} \times \mathbf{a} \neq 0$
Now, if $\mathbf{a}+x \mathbf{b}+y \mathbf{c}=0$
Let $x(\mathbf{a} \times \mathbf{b})=x y(\mathbf{b} \times \mathbf{c})=y(\mathbf{c} \times \mathbf{a})=\mathbf{p}$
$\Rightarrow \mathbf{a} \times \mathbf{b}=\frac{\mathbf{p}}{x}, \mathbf{b} \times \mathbf{c}=\frac{\mathbf{p}}{x y}$ and $\mathbf{c} \times \mathbf{a}=\frac{\mathbf{p}}{y}$
$\therefore \mathbf{a} \times \mathbf{b}+\mathbf{b} \times \mathbf{c}+\mathbf{c} \times \mathbf{a}=\frac{\mathbf{p}}{x}+\frac{\mathbf{p}}{x y}+\frac{\mathbf{p}}{y}$
$
=\mathbf{p}\left(\frac{x+y+1}{x y}\right)
$
$
\begin{aligned}
& =\left(\frac{x+y+1}{x y}\right) \times x y \times(\mathbf{b} \times \mathbf{c}) \\
& =(x+y+1)(\mathbf{b} \times \mathbf{c})
\end{aligned}
$
Comparing it with $6(\mathbf{b} \times \mathbf{c})$,
$
\begin{aligned}
& x+y+1=6 \\
& \therefore \quad x+y=5 \\
& \Rightarrow \quad x+y-5=0 \\
&
\end{aligned}
$
Hence, option (b) satisfies it