A symmetrical form of the line of intersection of the planes $$x = ay + b$$ and $$z = cy + d$$ is:

We are given two planes: $$x = ay + b$$ and $$z = cy + d$$. To find the symmetrical form of their line of intersection, we first rewrite the plane equations in standard form. The first plane is $$x - ay = b$$, and the second plane is $$-cy + z = d$$.

The normal vector to the first plane is $$\langle 1, -a, 0 \rangle$$ and to the second plane is $$\langle 0, -c, 1 \rangle$$. The direction vector of the line of intersection is perpendicular to both normals, so we compute their cross product:

$$$ \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -a & 0 \\ 0 & -c & 1 \\ \end{vmatrix} = \hat{i}[(-a)(1) - (0)(-c)] - \hat{j}[(1)(1) - (0)(0)] + \hat{k}[(1)(-c) - (-a)(0)] $$$

Simplifying each component:

$$$ \hat{i}[-a - 0] - \hat{j}[1 - 0] + \hat{k}[-c - 0] = -a\hat{i} - \hat{j} - c\hat{k} $$$

So, the direction vector is $$\langle -a, -1, -c \rangle$$, which can be simplified to $$\langle a, 1, c \rangle$$ by multiplying by $$-1$$. Thus, the direction ratios are $$a$$, $$1$$, and $$c$$.

Next, we find a point on the line by choosing a value for $$y$$. Let $$y = 1$$:

• From the first plane: $$x = a(1) + b = a + b$$
• From the second plane: $$z = c(1) + d = c + d$$

So, the point is $$(a + b, 1, c + d)$$.

The symmetrical form of a line is $$\frac{x - x_0}{l} = \frac{y - y_0}{m} = \frac{z - z_0}{n}$$, where $$(x_0, y_0, z_0)$$ is a point on the line and $$\langle l, m, n \rangle$$ is the direction vector. Substituting the point $$(a + b, 1, c + d)$$ and direction ratios $$\langle a, 1, c \rangle$$:

$$$ \frac{x - (a + b)}{a} = \frac{y - 1}{1} = \frac{z - (c + d)}{c} $$$

Rewriting the numerators:

$$$ \frac{x - a - b}{a} = \frac{y - 1}{1} = \frac{z - c - d}{c} $$$

Now, comparing with the options:

• Option A: $$\frac{x - b}{a} = \frac{y - 1}{1} = \frac{z - d}{c}$$ corresponds to the point $$(b, 1, d)$$, which does not satisfy the first plane unless $$a = 0$$ (since $$b = a(1) + b$$ implies $$a = 0$$), so it is incorrect.
• Option B: $$\frac{x - b - a}{a} = \frac{y - 1}{1} = \frac{z - d - c}{c}$$ matches exactly with our derived form.
• Option C: $$\frac{x - a}{b} = \frac{y - 0}{1} = \frac{z - c}{d}$$ has denominators $$b$$ and $$d$$ instead of $$a$$ and $$c$$, and the point $$(a, 0, c)$$ may not lie on the planes (e.g., $$a = a(0) + b$$ implies $$a = b$$, which is not necessarily true).
• Option D: $$\frac{x - b - a}{b} = \frac{y - 1}{0} = \frac{z - d - c}{d}$$ has a denominator of $$0$$ for $$y$$, which is invalid since the direction vector has a non-zero $$y$$-component ($$m = 1$$).

Therefore, only option B is correct.

Hence, the correct answer is Option B.