The vectors $$5\hat{i}+\hat{j}+\hat{k}, \hat{i}+5\hat{j}+\hat{k}$$ and $$\hat{i}+\hat{j}+5\hat{k}$$ are the three face diagonals of three faces having a common vertex in a parallelopiped. What is the volume of the parallelopiped?

Let the three given face diagonals meeting at a common vertex be represented as:

$$\vec{d_1} = 5\hat{i} + \hat{j} + \hat{k}$$

$$\vec{d_2} = \hat{i} + 5\hat{j} + \hat{k}$$

$$\vec{d_3} = \hat{i} + \hat{j} + 5\hat{k}$$

If $$\vec{a}$$, $$\vec{b}$$, and $$\vec{c}$$ are the three coterminous edges of the parallelepiped meeting at that vertex, the face diagonals can be expressed as:

$$\vec{d_1} = \vec{a} + \vec{b}$$

$$\vec{d_2} = \vec{b} + \vec{c}$$

$$\vec{d_3} = \vec{c} + \vec{a}$$

The scalar triple product of the face diagonals satisfies the standard relation:

$$[\vec{d_1} \ \vec{d_2} \ \vec{d_3}] = 2 [\vec{a} \ \vec{b} \ \vec{c}]$$

Thus, the volume $$V$$ of the parallelepiped is equal to half of the scalar triple product of the face diagonals:

$$V = \frac{1}{2} \left\vert{} \begin{matrix} 5 & 1 & 1 \\ 1 & 5 & 1 \\ 1 & 1 & 5 \end{matrix} \right\vert{}$$

Evaluating the determinant by expanding along the first row:

$$\text{Determinant} = 5(5 \times 5 - 1 \times 1) - 1(1 \times 5 - 1 \times 1) + 1(1 \times 1 - 5 \times 1)$$

$$\text{Determinant} = 5(24) - 1(4) + 1(-4)$$

$$\text{Determinant} = 120 - 4 - 4 = 112$$

Substituting this value into the volume expression:

$$V = \frac{112}{2} = 56$$

The correct option is B.