Explanation

(b)
The equation of the line through the centre $\widehat{j} + 2\widehat{k}$ and normal to the given plane is
$\overrightarrow{r} = \widehat{j} + 2\widehat{k} + \lambda(\widehat{i} + 2\widehat{j} + 2\widehat{k})$ (i)
This meets the plane for which
$\left\lbrack \widehat{j} + 2\widehat{k} + \lambda\left( \widehat{i} + 2\widehat{j} + 2\widehat{k} \right) \right\rbrack \bullet (\widehat{i} + 2\widehat{j} + 2\widehat{k}) = 15$
$\Rightarrow 6 + 9\lambda = 15\ \Rightarrow \ \lambda = 1$
Putting in (i), we get
$\overrightarrow{r} = \widehat{j} + 2\widehat{k} + \left( \widehat{i} + 2\widehat{j} + 2\widehat{k} \right) = \widehat{i} + 3\widehat{j} + 4\widehat{k}$
Hence, centre is (1, 3, 4)