Let $$\overrightarrow{u},\overrightarrow{v}$$ and $$\overrightarrow{w}$$ be vectors in three-dimensional space, where $$\overrightarrow{u}$$ and $$\overrightarrow{v}$$ re unit vectors which are not perpendicular to each other and
$$\overrightarrow{u}.\overrightarrow{w} = 1, \overrightarrow{v}.\overrightarrow{w} = 1, \overrightarrow{w}\overrightarrow{w} = 4$$
If the volume of the parallelopiped, whose adjacent sides are represented by the vectors $$\overrightarrow{u},\overrightarrow{v}$$ and $$\overrightarrow{w}$$, is $$\sqrt{2}$$, then the value of $$\mid 3\overrightarrow{u} + 5 \overrightarrow{v} \mid$$ is ___ .

Let the unit vectors $$\overrightarrow{u},\overrightarrow{v}$$ enclose an angle $$\theta$$ so that
$$\overrightarrow{u}\cdot\overrightarrow{v}=\cos\theta,\qquad |\overrightarrow{u}|=|\overrightarrow{v}|=1$$

Create an orthonormal frame with
$$\overrightarrow{u}=(1,0,0),\qquad \overrightarrow{n}=(0,1,0)$$
and write $$\overrightarrow{v}$$ in that frame:
$$\overrightarrow{v}=(\cos\theta,\;\sin\theta,\;0).$$

Write $$\overrightarrow{w}=(a,b,c).$$ The given dot-product conditions give

$$a=\overrightarrow{u}\cdot\overrightarrow{w}=1\qquad\text{and}\qquad \overrightarrow{v}\cdot\overrightarrow{w}=a\cos\theta+b\sin\theta=1.$$

With $$a=1$$, the second relation yields
$$b=\dfrac{1-\cos\theta}{\sin\theta}.$$

The length of $$\overrightarrow{w}$$ is $$|\overrightarrow{w}|=\sqrt{4}=2,$$ so

$$1^{2}+b^{2}+c^{2}=4\;\;\Longrightarrow\;\;c^{2}=3-b^{2}.$$ Let $$\cos\theta=k,$$ hence $$\sin\theta=\sqrt{1-k^{2}}$$ and

$$b^{2}=\left(\dfrac{1-k}{\sqrt{1-k^{2}}}\right)^{2} =\dfrac{(1-k)^{2}}{1-k^{2}} =\dfrac{1-k}{1+k}.$$ Thus
$$c^{2}=3-\dfrac{1-k}{1+k} =\dfrac{2+4k}{1+k}=2\dfrac{1+2k}{1+k}.$$

The volume of the parallelepiped is
$$V=|\overrightarrow{u}\cdot(\overrightarrow{v}\times\overrightarrow{w})|=\sqrt{2}.$$
Because $$\overrightarrow{u}=(1,0,0),$$ the x-component of $$\overrightarrow{v}\times\overrightarrow{w}$$ alone matters:

$$\overrightarrow{v}\times\overrightarrow{w}= \bigl(\sin\theta\,c,\;-\cos\theta\,c,\;\cos\theta\,b-\sin\theta\bigr)$$ so $$V=|\sin\theta\,c|=\sqrt{2}.$$

Square both sides:

$$\sin^{2}\theta\,c^{2}=2 \;\;\Longrightarrow\;\; (1-k^{2})\,c^{2}=2.$$

Substitute the expression for $$c^{2}$$:

$$(1-k^{2})\cdot 2\dfrac{1+2k}{1+k}=2.$$

Simplify using $$1-k^{2}=(1-k)(1+k):$$

$$(1-k)(1+k)\dfrac{1+2k}{1+k}=1 \;\;\Longrightarrow\;\; (1-k)(1+2k)=1.$$

Expand and rearrange:

$$1+k-2k^{2}=1\;\;\Longrightarrow\;\;k-2k^{2}=0 \;\;\Longrightarrow\;\;k(1-2k)=0.$$

Since $$\overrightarrow{u}$$ and $$\overrightarrow{v}$$ are not perpendicular, $$k\neq0,$$ hence $$1-2k=0$$ and

$$\cos\theta=k=\dfrac12\qquad\Longrightarrow\qquad\theta=60^{\circ}.$$

Finally, compute $$\bigl|\,3\overrightarrow{u}+5\overrightarrow{v}\bigr|$$ using the cosine rule:

$$\begin{aligned} \bigl|\,3\overrightarrow{u}+5\overrightarrow{v}\bigr|^{2} &=9|\overrightarrow{u}|^{2}+25|\overrightarrow{v}|^{2} +2\cdot3\cdot5(\overrightarrow{u}\cdot\overrightarrow{v})\\ &=9+25+30\cos\theta\\ &=34+30\left(\dfrac12\right)\\ &=34+15=49. \end{aligned}$$

Therefore
$$\bigl|\,3\overrightarrow{u}+5\overrightarrow{v}\bigr|=\sqrt{49}=7.$$

Answer: 7