An angle between the plane, $$x + y + z = 5$$ and the line of intersection of the planes, $$3x + 4y + z - 1 = 0$$ and $$5x + 8y + 2z + 14 = 0$$, is:

We are asked to find the acute angle between the plane

$$x + y + z = 5$$

and the line which is obtained as the intersection of the two planes

$$3x + 4y + z - 1 = 0 \quad\text{and}\quad 5x + 8y + 2z + 14 = 0.$$

For any plane, the vector perpendicular to it is called its normal. So we first list the normals of the three planes involved:

For $$x + y + z - 5 = 0,$$ the normal vector is $$\mathbf{n}_0 = (1,\,1,\,1).$$

For $$3x + 4y + z - 1 = 0,$$ the normal vector is $$\mathbf{n}_1 = (3,\,4,\,1).$$

For $$5x + 8y + 2z + 14 = 0,$$ the normal vector is $$\mathbf{n}_2 = (5,\,8,\,2).$$

Now, the required line is the intersection of the second and third planes. A direction vector of this line is perpendicular to both $$\mathbf{n}_1$$ and $$\mathbf{n}_2$$. Such a vector is given by their cross-product:

$$\mathbf{d} \;=\; \mathbf{n}_1 \times \mathbf{n}_2.$$

We compute the cross-product term by term:

$$ \mathbf{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 4 & 1 \\ 5 & 8 & 2 \end{vmatrix} = \mathbf{i}(4\cdot2 - 1\cdot8)\;-\;\mathbf{j}(3\cdot2 - 1\cdot5)\;+\;\mathbf{k}(3\cdot8 - 4\cdot5). $$

Simplifying each component:

$$ \mathbf{i}(8 - 8) = \mathbf{i}\,0, \qquad -\mathbf{j}(6 - 5) = -\mathbf{j}\,1, \qquad \mathbf{k}(24 - 20) = \mathbf{k}\,4. $$

So

$$\mathbf{d} = (0,\,-1,\,4).$$

Next we find the angle between the line having direction $$\mathbf{d}$$ and the plane

First recall the relation connecting the angle between a line and a plane with the angle between the line and the plane’s normal. If $$\phi$$ is the angle between the direction vector of the line and the normal vector of the plane, then

$$\theta = 90^\circ - \phi,$$

which implies

$$\sin\theta = \cos\phi.$$

And we know the cosine of $$\phi$$ from the scalar (dot) product formula:

$$\cos\phi = \frac{|\mathbf{d}\cdot\mathbf{n}_0|}{\|\mathbf{d}\|\,\|\mathbf{n}_0\|}.$$

Therefore directly,

$$\sin\theta = \frac{|\mathbf{d}\cdot\mathbf{n}_0|}{\|\mathbf{d}\|\,\|\mathbf{n}_0\|}.$$

We now evaluate each quantity in this expression.

Dot product.

$$\mathbf{d}\cdot\mathbf{n}_0 = (0)(1) + (-1)(1) + (4)(1) = -1 + 4 = 3.$$

Taking absolute value gives $$|\mathbf{d}\cdot\mathbf{n}_0| = 3.$$

Magnitude of $$\mathbf{d}$$.

$$\|\mathbf{d}\| = \sqrt{\,0^{2} + (-1)^{2} + 4^{2}\,} = \sqrt{0 + 1 + 16} = \sqrt{17}.$$

Magnitude of $$\mathbf{n}_0$$.

$$\|\mathbf{n}_0\| = \sqrt{\,1^{2} + 1^{2} + 1^{2}\,} = \sqrt{3}.$$

Substituting these three numerical values into the formula for $$\sin\theta$$ we get

$$ \sin\theta = \dfrac{3}{\sqrt{17}\,\sqrt{3}} = \dfrac{3}{\sqrt{51}}. $$

Notice that $$\sqrt{51} = \sqrt{17}\sqrt{3},$$ so we can cancel one factor of $$\sqrt{3}$$ in the numerator and denominator:

$$ \dfrac{3}{\sqrt{17}\,\sqrt{3}} = \dfrac{\sqrt{3}\,\sqrt{3}}{\sqrt{17}\,\sqrt{3}} = \dfrac{\sqrt{3}}{\sqrt{17}} = \sqrt{\frac{3}{17}}. $$

Thus we have obtained

$$\sin\theta = \sqrt{\frac{3}{17}}.$$

Taking the inverse sine (principal value) gives

$$\theta = \sin^{-1}\!\left(\sqrt{\frac{3}{17}}\right).$$

This matches Option D.

Hence, the correct answer is Option D.