Let $$\vec{a}, \vec{b}, \vec{c}$$ be three non-coplanar vectors such that $$\vec{a} \times \vec{b} = 4\vec{c}$$, $$\vec{b} \times \vec{c} = 9\vec{a}$$ and $$\vec{c} \times \vec{a} = \alpha\vec{b}$$, $$\alpha > 0$$. If $$|\vec{a}| + |\vec{b}| + |\vec{c}| = {36}$$, then $$\alpha$$ is equal to _______.

Given,

$$\vec a\times\vec b=4\vec c$$

$$\vec b\times\vec c=9\vec a$$

$$\vec c\times\vec a=\alpha\vec b,\qquad \alpha>0$$

Now,

$$\vec a\times\vec b$$

is perpendicular to both

$$\vec a\quad \text{and}\quad \vec b$$

Since

$$\vec a\times\vec b=4\vec c,$$

we get

$$\vec c\perp\vec a,\qquad \vec c\perp\vec b$$

Similarly,

$$\vec b\times\vec c=9\vec a$$

implies

$$\vec a\perp\vec b,\qquad \vec a\perp\vec c$$

Hence, all three vectors are mutually perpendicular.

Therefore, angle between every pair is

$$90^\circ$$

So,

$$\sin90^\circ=1$$

Taking magnitudes in the given equations,

$$|\vec a||\vec b|=4|\vec c|\qquad\cdots(1)$$

$$|\vec b||\vec c|=9|\vec a|\qquad\cdots(2)$$

$$|\vec c||\vec a|=\alpha|\vec b|\qquad\cdots(3)$$

From (1) and (2),

$$\frac{|\vec a|}{|\vec c|}=\frac{4|\vec c|}{9|\vec a|}$$

$$9|\vec a|^2=4|\vec c|^2$$

$$3|\vec a|=2|\vec c|$$

Let

$$|\vec a|=2k,\qquad |\vec c|=3k$$

Using (1),

$$2k\cdot|\vec b|=4(3k)$$

$$|\vec b|=6$$

Now use (2),

$$6(3k)=9(2k)$$

which is satisfied.

Using (3),

$$3k(2k)=\alpha(6)$$

$$6k^2=6\alpha$$

$$k^2=\alpha$$

Hence,

$$|\vec a|=2\sqrt\alpha,\qquad |\vec c|=3\sqrt\alpha$$

Also,

$$|\vec b|=6$$

Given,

$$|\vec a|+|\vec b|+|\vec c|=36$$

$$2\sqrt\alpha+6+3\sqrt\alpha=36$$

$$5\sqrt\alpha=30$$

$$\sqrt\alpha=6$$

Therefore,

$$\alpha=36$$

Hence, the required value is

$$\boxed{36}$$.