Given,

$$\overrightarrow a \times \overrightarrow b = 4\,.\,\overrightarrow c $$ ..... (i)

$$\overrightarrow b \times \overrightarrow c = 9\,.\,\overrightarrow a $$ ..... (ii)

$$\overrightarrow c \times \overrightarrow a = \alpha \,.\,\overrightarrow b $$ .... (iii)

Taking dot products with $$\overrightarrow c ,\overrightarrow a ,\overrightarrow b $$ we get

$$\overrightarrow a \,.\,\overrightarrow b = \overrightarrow b \,.\,\overrightarrow c = \overrightarrow c \,.\,\overrightarrow a = 0$$

Hence,

(i) $$ \Rightarrow |\overrightarrow a |\,.\,|\overrightarrow b | = 4\,.\,|\overrightarrow c |$$ ..... (iv)

(ii) $$ \Rightarrow |\overrightarrow b |\,.\,|\overrightarrow c | = 9\,.\,|\overrightarrow a |$$ ..... (v)

(iii) $$ \Rightarrow |\overrightarrow c |\,.\,|\overrightarrow a | = \alpha \,.\,|\overrightarrow b |$$ .... (vi)

Multiplying (iv), (v) and (vi)

$$ \Rightarrow |\overrightarrow a |\,.\,|\overrightarrow b |\,.\,|\overrightarrow c | = 36\alpha $$ ..... (vii)

Dividing (vii) by (iv) $$ \Rightarrow |\overrightarrow c {|^2} = 9\alpha \Rightarrow |\overrightarrow c | = 3\sqrt \alpha $$ ..... (viii)

Dividing (vii) by (v) $$ \Rightarrow |\overrightarrow a {|^2} = 4\alpha \Rightarrow |\overrightarrow a | = 2\sqrt \alpha $$

Dividing (viii) by (vi) $$ \Rightarrow |\overrightarrow b {|^2} = 36 \Rightarrow |\overrightarrow b | = 6$$

Now, as given, $$3\sqrt \alpha + 2\sqrt \alpha + 6 = {1 \over {36}} \Rightarrow \sqrt \alpha = {{ - 43} \over {36}}$$