Explanation

(b)
Let $\overrightarrow{\mathbf{r}} = x\widehat{\dot{\mathbf{i}}} + y\widehat{\dot{\mathbf{j}}} + z\widehat{\mathbf{k}}$
Given, $\overrightarrow{\mathbf{r}} \times \overrightarrow{\mathbf{b}} = \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{b}}$
$\Longrightarrow \left( x\widehat{\dot{\mathbf{i}}} + y\widehat{\dot{\mathbf{j}}} + z\widehat{\mathbf{k}} \right) \times \left( \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}} \right)$
$= \left( 4\widehat{\dot{\mathbf{i}}} - 3\widehat{\dot{\mathbf{j}}} + 7\widehat{\mathbf{k}} \right) \times \left( \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}} \right)$
$\Longrightarrow (y - z)\widehat{\dot{\mathbf{i}}} - (x - z)\widehat{\dot{\mathbf{j}}} + (x - y)\widehat{\mathbf{k}}$
$= - 10\widehat{\dot{\mathbf{i}}} + 3\widehat{\dot{\mathbf{j}}} + 7\widehat{\mathbf{k}}$
$\Longrightarrow y - z = - 10, - (x - z) = 3,x - y = 7$
$\Longrightarrow y - z = - 10, - x + z = 3,x - y = 7$ ….(i)
and $\overrightarrow{\mathbf{r}} \bullet \overrightarrow{\mathbf{a}} = 0$
$\Longrightarrow \left( x\widehat{\dot{\mathbf{i}}} + y\widehat{\dot{\mathbf{j}}} + z\widehat{\mathbf{k}} \right) \bullet \left( 2\widehat{\dot{\mathbf{i}}} + \widehat{\mathbf{k}} \right)$
$\Longrightarrow 2x + z = 0\ \ \$……(ii)
From Eqs. (i) and (ii), we get
$x = - 1,y = - 8,z = 2$
$\therefore\ \overrightarrow{\mathbf{r}} \bullet \overrightarrow{\mathbf{b}} = \left( - \widehat{\dot{\mathbf{i}}} - 8\widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}} \right) \bullet \left( \widehat{\dot{\mathbf{i}}} + \widehat{\dot{\mathbf{j}}} + \widehat{\mathbf{k}} \right)$
$= - 1 - 8 + 2$
$= - 7$