Let the shortest distance between the lines $$\dfrac{x - 3}{3} = \dfrac{y - \alpha}{-1} = \dfrac{z - 3}{1}$$ and $$\dfrac{x + 3}{-3} = \dfrac{y + 7}{2} = \dfrac{z - \beta}{4}$$ be $$3\sqrt{30}$$. Then the positive value of $$5\alpha + \beta$$ is

The first line can be written in vector form as
$$\mathbf{r}= \mathbf{a}_1 + t\,\mathbf{b}_1,$$
where a point on the line is $$\mathbf{a}_1 = (3,\; \alpha,\; 3)$$ and the direction vector is $$\mathbf{b}_1 = \langle 3,\; -1,\; 1\rangle.$$

The second line is
$$\mathbf{r}= \mathbf{a}_2 + s\,\mathbf{b}_2,$$
with $$\mathbf{a}_2 = (-3,\; -7,\; \beta)$$ and $$\mathbf{b}_2 = \langle -3,\; 2,\; 4\rangle.$$

For two skew lines the shortest distance is given by
$$D = \frac{\left| (\mathbf{a}_2-\mathbf{a}_1)\cdot(\mathbf{b}_1 \times \mathbf{b}_2) \right|}{\left| \mathbf{b}_1 \times \mathbf{b}_2 \right|}. \qquad -(1)$$

Compute the cross product $$\mathbf{b}_1 \times \mathbf{b}_2$$:

$$ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ 3 & -1 & 1\\ -3 & 2 & 4 \end{vmatrix} = \langle (-1)(4)-1(2),\; -(3\cdot 4-1\cdot(-3)),\; 3\cdot 2-(-1)(-3) \rangle = \langle -6,\; -15,\; 3\rangle. $$

Thus $$\mathbf{b}_1 \times \mathbf{b}_2 = \langle -6,\; -15,\; 3\rangle = -3\langle 2,\; 5,\; -1\rangle.$$

Magnitude of this vector:
$$\left| \mathbf{b}_1 \times \mathbf{b}_2 \right| = 3\sqrt{2^{2}+5^{2}+(-1)^{2}} = 3\sqrt{30}. \qquad -(2)$$

Vector joining the two given points:
$$(\mathbf{a}_2-\mathbf{a}_1) = \langle -3-3,\; -7-\alpha,\; \beta-3\rangle = \langle -6,\; -7-\alpha,\; \beta-3\rangle.$

Now evaluate the scalar triple product in the numerator of (1):

$$ (\mathbf{a}_2-\mathbf{a}_1)$$\cdot$$(\mathbf{b}_1 $$\times$$ \mathbf{b}_2) = \langle -6,\; -7-$$\alpha$$,\; $$\beta$$-3\rangle $$\cdot$$ \langle -6,\; -15,\; 3\rangle $$ $$ = (-6)(-6) + (-7-$$\alpha$$)(-15) + ($$\beta$$-3)(3) $$ $$ = 36 + 15(7+$$\alpha$$) + 3($$\beta$$-3) $$ $$ = 36 + 105 + 15$$\alpha$$ + 3$$\beta$$ - 9 $$ $$ = 132 + 15$$\alpha$$ + 3$$\beta$$. \qquad -(3) $$

Given shortest distance $$D = 3$$\sqrt{30}$$,$$ substitute (2) and (3) into (1):

$$ $$\frac{\left| 132 + 15\alpha + 3\beta \right|}{3\sqrt{30}$$} = 3$$\sqrt{30}$$. $$

Multiply both sides by $$3$$\sqrt{30}$$:$$
$$$$\left$$| 132 + 15$$\alpha$$ + 3$$\beta$$ $$\right$$| = 9 $$\times$$ 30 = 270. \qquad -(4)$$

Equation (4) gives two possibilities:
$$132 + 15$$\alpha$$ + 3$$\beta$$ = 270 \quad$$\text{or}$$\quad 132 + 15$$\alpha$$ + 3$$\beta$$ = -270.$$ Divide each by $$3$$:

Case 1: $$5$$\alpha + \beta$$ = 46.$$
Case 2: $$5$$\alpha + \beta$$ = -134.$$

The problem asks for the positive value of $$5$$\alpha + \beta$$$$, hence we choose
Case 1: $$5$$\alpha + \beta$$ = 46.$$

Therefore, the required positive value is $$\mathbf{46}$$, which corresponds to Option B.