If for a >0, the feet of perpendiculars from

C lies on plane $\Rightarrow-m a-n=0 \Rightarrow \frac{m}{n}=-\frac{1}{a} \ldots . .(1)$

$\overrightarrow{\mathrm{CA}} \| \hat{\mathrm{i}}+\mathrm{m} \hat{\mathrm{j}}+\mathrm{nk}$

$\frac{\mathrm{a}-0}{l}=\frac{-\mathrm{a}}{\mathrm{m}}=\frac{4}{\mathrm{n}} \Rightarrow \frac{\mathrm{m}}{\mathrm{n}}=-\frac{\mathrm{a}}{4}$......(2)

From (1) & (2)

$-\frac{1}{a}=\frac{-a}{4} \Rightarrow a^{2}=4 \Rightarrow a=2 \quad($ since $a>0)$

From (2) $\frac{\mathrm{m}}{\mathrm{n}}=\frac{-1}{2}$

Let $m=-t \Rightarrow n=2 t$

$\frac{2}{l}=\frac{-2}{-\mathrm{t}} \Rightarrow l=\mathrm{t}$

So plane : $\mathrm{t}(\mathrm{x}-\mathrm{y}+2 \mathrm{z})=0$

$\mathrm{BD}=\frac{6}{\sqrt{6}}=\sqrt{6} \quad \mathrm{C} \cong(0,-2,-1)$

$\mathrm{CD}=\sqrt{\mathrm{BC}^{2}-\mathrm{BD}^{2}}$

$=\sqrt{\left(0^{2}+6^{2}+6^{2}\right)-(\sqrt{6})^{2}}$

$=\sqrt{66}$