Explanation

(c)
The equation of a line passing through the centre $(\widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}})$ and normal to the given plane is
$\overrightarrow{\mathbf{r}} = \widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}} + \lambda(\widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}}$) …..(i)
This meets the plane at a point for which we must have
$\left\lbrack \left( \widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}} \right) + \lambda\left( \widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}} \right) \right\rbrack.\left( \widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} + 2\widehat{\mathbf{k}} \right) = 15$
$\Longrightarrow 6 + \lambda(9) = 15 \Longrightarrow \lambda = 1$
$\therefore$ From Eq. (i),
$\overrightarrow{\mathbf{r}} = \widehat{\dot{\mathbf{i}}} + 3\widehat{\dot{\mathbf{j}}} + 4\widehat{\mathbf{k}}$
$\therefore$ Coordinates of the centre of the circle are (1,3,4)