Explanation

(c)
Given that, $\overrightarrow{\mathbf{a}}\mathbf{=}\widehat{\mathbf{i}}\mathbf{+}\widehat{\mathbf{j}}\ and\ \overrightarrow{\mathbf{b}}\mathbf{=}2\widehat{\mathbf{i}}\mathbf{-}\widehat{\mathbf{k}}$
Let $\overrightarrow{\mathbf{r}}\mathbf{=}x\widehat{\mathbf{i}} + y\widehat{\mathbf{j}}\mathcal{+ z}\widehat{\mathbf{k}}$, then
$\overrightarrow{\mathbf{r}}\mathbf{\times}\overrightarrow{\mathbf{a}}\mathbf{=}\overrightarrow{\mathbf{b}}\mathbf{\times}\overrightarrow{\mathbf{a}}\ \Rightarrow \left( \mathbf{\ }\overrightarrow{\mathbf{r}}\mathbf{-}\overrightarrow{\mathbf{b}} \right)\mathbf{\times}\overrightarrow{\mathbf{a}}\mathbf{=}\overrightarrow{\mathbf{0}}$
Now, $\overrightarrow{\mathbf{r}}\mathbf{-}\overrightarrow{\mathbf{b}} = \left( x\widehat{\mathbf{i}} + y\widehat{\mathbf{j}}\mathcal{+ z}\widehat{\mathbf{k}} \right) - (2\widehat{\mathbf{i}} - \widehat{\mathbf{k}})$
$= (x - 2)\widehat{\mathbf{i}} + y\widehat{\mathbf{j}}\mathcal{+ (z} + 1)\widehat{\mathbf{k}}$
$\therefore\ \left( \mathbf{\ }\overrightarrow{\mathbf{r}}\mathbf{-}\overrightarrow{\mathbf{b}} \right)\mathbf{\times}\overrightarrow{\mathbf{a}} = \left| \begin{matrix}
\widehat{\mathbf{i}} & \widehat{\mathbf{j}} & \widehat{\mathbf{k}} \\
x - 2 & y & \mathcal{z} + 1 \\
1 & 1 & 0
\end{matrix} \right| = \overrightarrow{\mathbf{0}}$
$\Rightarrow \ - \left( \mathcal{z} + 1 \right)\widehat{\mathbf{i}} + \left( \mathcal{z} + 1 \right)\widehat{\mathbf{j}} + (x - 2 - y)\widehat{\mathbf{k}} = \overrightarrow{\mathbf{0}}$
On equating the coefficient of $\widehat{i},\widehat{j}\ and\ \widehat{k}$, we get
$\mathcal{z} = - 1,\ x - y = 2$ ….(i)
Now, $\overrightarrow{\mathbf{r}}\mathbf{\times}\overrightarrow{\mathbf{b}}\mathbf{=}\overrightarrow{\mathbf{a}}\mathbf{\times}\overrightarrow{\mathbf{b}}\mathbf{\ }\mathbf{\Rightarrow \ }\left( \mathbf{\ }\overrightarrow{\mathbf{r}}\mathbf{-}\overrightarrow{\mathbf{a}} \right)\mathbf{\times}\overrightarrow{\mathbf{b}}$=$\overrightarrow{\mathbf{0}}$
And $\overrightarrow{\mathbf{r}}\mathbf{-}\overrightarrow{\mathbf{a}} = (x - 1)\widehat{\mathbf{i}} + (y - 1)\widehat{\mathbf{j}} + \mathcal{z}\widehat{\mathbf{k}}$
$\therefore\left( \overrightarrow{\mathbf{r}}\mathbf{-}\overrightarrow{\mathbf{a}} \right)\mathbf{\times}\overrightarrow{\mathbf{b}} = \left| \begin{matrix}
\widehat{\mathbf{i}} & \widehat{\mathbf{j}} & \widehat{\mathbf{k}} \\
x - 1 & y - 1 & \mathcal{z} \\
2 & 0 & - 1
\end{matrix} \right| = \overrightarrow{\mathbf{0}}\$
$\Rightarrow ( - y + 1)\widehat{\mathbf{i}}\mathbf{-}\widehat{\mathbf{j}}\left( - x + 1 - 2\mathcal{z} \right) + ( - 2y + 2)\widehat{\mathbf{k}}\mathbf{=}\overrightarrow{\mathbf{0}}\$
$\Rightarrow y = 1,\ x + 2\mathcal{z} = 1$ ...(ii)
From Eqs. (i) and (ii), we get
$x = 3,\ y = 1\ \mathcal{z} = - 1$
$\therefore\overrightarrow{\mathbf{r}} = \ 3\widehat{\mathbf{i}} + \widehat{\mathbf{j}} - \widehat{\mathbf{k}}$