Explanation

(c)
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Equation of a line $AB$ is $\frac{x - 1}{1} = \frac{y}{- 2} = \frac{\mathcal{z}}{0} = \lambda$
Now $AB\bot OC \Rightarrow \ 1(\lambda + 1) + ( - 2\lambda)( - 2) = 0\ \Rightarrow 5\lambda = - 1\ \Rightarrow \ \lambda = - \frac{1}{5}$
$\Rightarrow C$ is $\left( \frac{4}{5},\frac{2}{5},0 \right)$. Now
$x_{1}^{2} + \left( y_{1} - 2 \right)^{2} + \mathcal{z}_{1}^{2} = 4$ (i)
and $\left( x_{1} - 1 \right)^{2} + y_{1}^{2} + \mathcal{z}_{1}^{2} = 1$ (ii)
Now $OC\bot CD$
$\Rightarrow \ \left( x_{1} - \frac{4}{5} \right)\frac{4}{5} + \left( y_{1} - \frac{2}{5} \right)\frac{2}{5} + \left( \mathcal{z}_{1} - 0 \right)0 = 0$ (iii)
From (i) and (ii), we get
$- 4y_{1} + 2x_{1} = 0\ \Rightarrow \ \ x_{1} = 2y_{1}$
From (iii), putting $x_{1} = 2y_{1} \Rightarrow 2y_{1} = \frac{4}{5}\ \Rightarrow y_{1} = \frac{2}{5}\ \Rightarrow x_{1} = \frac{4}{5}$. Putting this value of $x_{1}$ and $y_{1}$ in (i), we get
$\mathcal{z}_{1} = \pm \frac{2}{\sqrt{5}}$