Step-by-step Solution:

1. Define the Integral and Apply King's Rule

   Let the given integral be denoted by $I$. $I = \int\limits_0^{\pi /2} {{{3\sqrt {\cos \theta } } \over {{{(\sqrt {\cos \theta } + \sqrt {\sin \theta } )}^5}}}} d\theta \quad \quad ...(1)$ We use the property of definite integrals, often called King's Rule: $\int_0^a f(x) dx = \int_0^a f(a-x) dx$. Here, $a = \pi/2$. Applying this property to $I$: $I = \int\limits_0^{\pi /2} {{{3\sqrt {\cos (\pi/2 - \theta) } } \over {{{(\sqrt {\cos (\pi/2 - \theta) } + \sqrt {\sin (\pi/2 - \theta) } )}^5}}}} d\theta$ Using the identities $\cos(\pi/2 - \theta) = \sin \theta$ and $\sin(\pi/2 - \theta) = \cos \theta$, we get: $I = \int\limits_0^{\pi /2} {{{3\sqrt {\sin \theta } } \over {{{(\sqrt {\sin \theta } + \sqrt {\cos \theta } )}^5}}}} d\theta \quad \quad ...(2)$

2. Combine the two expressions for the Integral

   Adding equation (1) and (2), we get: $2I = \int\limits_0^{\pi /2} {{{3\sqrt {\cos \theta } } \over {{{(\sqrt {\cos \theta } + \sqrt {\sin \theta } )}^5}}}} d\theta + \int\limits_0^{\pi /2} {{{3\sqrt {\sin \theta } } \over {{{(\sqrt {\sin \theta } + \sqrt {\cos \theta } )}^5}}}} d\theta$ $2I = \int\limits_0^{\pi /2} {{{3(\sqrt {\cos \theta } + \sqrt {\sin \theta }) } \over {{{(\sqrt {\cos \theta } + \sqrt {\sin \theta } )}^5}}}} d\theta$ This simplifies to: $2I = \int\limits_0^{\pi /2} {{{3 } \over {{{(\sqrt {\cos \theta } + \sqrt {\sin \theta } )}^4}}}} d\theta$

3. Simplify the Integrand for Substitution

   To simplify the integrand, we can factor out $\sqrt{\cos \theta}$ from the denominator: $(\sqrt {\cos \theta } + \sqrt {\sin \theta } )^4 = \left[ \sqrt{\cos\theta} \left(1 + \sqrt{\frac{\sin\theta}{\cos\theta}}\right) \right]^4 = (\sqrt{\cos\theta})^4 (1 + \sqrt{\tan\theta})^4 = \cos^2\theta (1 + \sqrt{\tan\theta})^4$ Substituting this back into the expression for $2I$: $2I = \int\limits_0^{\pi /2} {{{3 } \over {\cos^2\theta (1 + \sqrt{\tan\theta})^4}}} d\theta = \int\limits_0^{\pi /2} {{{3\sec^2\theta } \over {(1 + \sqrt{\tan\theta})^4}}} d\theta$

4. Perform a Substitution

   Let $t = \tan\theta$. Then $dt = \sec^2\theta \, d\theta$. We must also change the limits of integration:

   • When $\theta = 0$, $t = \tan(0) = 0$.
   • When $\theta = \pi/2$, $t \to \infty$. The integral becomes: $2I = \int\limits_0^{\infty} {{{3 } \over {(1 + \sqrt{t})^4}}} dt$

5. Perform a Second Substitution

   Let $u = 1 + \sqrt{t}$. Then $\sqrt{t} = u - 1$, so $t = (u-1)^2$. Differentiating with respect to $u$, we get $dt = 2(u-1)du$. The new limits of integration are:

   • When $t = 0$, $u = 1 + \sqrt{0} = 1$.
   • When $t \to \infty$, $u \to \infty$. Substituting these into the integral for $2I$: $2I = \int\limits_1^{\infty} {{{3 } \over {u^4}}} \cdot 2(u-1)du = 6 \int\limits_1^{\infty} {{{u-1 } \over {u^4}}} du$

6. Evaluate the Final Integral

   We can now evaluate the integral: $2I = 6 \int\limits_1^{\infty} \left(\frac{u}{u^4} - \frac{1}{u^4}\right) du = 6 \int\limits_1^{\infty} (u^{-3} - u^{-4}) du$ $2I = 6 \left[ \frac{u^{-2}}{-2} - \frac{u^{-3}}{-3} \right]_1^{\infty} = 6 \left[ -\frac{1}{2u^2} + \frac{1}{3u^3} \right]_1^{\infty}$ Evaluating at the limits: $2I = 6 \left[ \left( \lim_{u\to\infty} \left(-\frac{1}{2u^2} + \frac{1}{3u^3}\right) \right) - \left( -\frac{1}{2(1)^2} + \frac{1}{3(1)^3} \right) \right]$ $2I = 6 \left[ (0 - 0) - \left( -\frac{1}{2} + \frac{1}{3} \right) \right]$ $2I = 6 \left[ - \left( \frac{-3+2}{6} \right) \right] = 6 \left[ - \left( -\frac{1}{6} \right) \right] = 6 \left( \frac{1}{6} \right) = 1$

7. Calculate the value of I

   Since $2I = 1$, the value of the integral is: $I = \frac{1}{2} = 0.5$