Show that the lines $\frac{x-a+d}{\alpha-\delta}=\frac{y-a}{\alpha}=\frac{z-a-d}{\alpha+\delta}$ and $\frac{x-b+c}{\beta-\gamma}=\frac{y-b}{\beta}=\frac{z-b-c}{\beta+\gamma}$ are coplanar.

(A) Two lines $\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2}$ are coplanar if $\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$.
For the given lines,we have:
$(x_1, y_1, z_1) = (a-d, a, a+d)$ and $(a_1, b_1, c_1) = (\alpha-\delta, \alpha, \alpha+\delta)$.
$(x_2, y_2, z_2) = (b-c, b, b+c)$ and $(a_2, b_2, c_2) = (\beta-\gamma, \beta, \beta+\gamma)$.
Now,consider the determinant:
$\Delta = \begin{vmatrix} b-c-a+d & b-a & b+c-a-d \\ \alpha-\delta & \alpha & \alpha+\delta \\ \beta-\gamma & \beta & \beta+\gamma \end{vmatrix}$.
Applying the column operation $C_1 \to C_1 + C_3$:
$\Delta = \begin{vmatrix} (b-c-a+d) + (b+c-a-d) & b-a & b+c-a-d \\ (\alpha-\delta) + (\alpha+\delta) & \alpha & \alpha+\delta \\ (\beta-\gamma) + (\beta+\gamma) & \beta & \beta+\gamma \end{vmatrix} = \begin{vmatrix} 2(b-a) & b-a & b+c-a-d \\ 2\alpha & \alpha & \alpha+\delta \\ 2\beta & \beta & \beta+\gamma \end{vmatrix}$.
Taking $2$ as a common factor from the first column:
$\Delta = 2 \begin{vmatrix} b-a & b-a & b+c-a-d \\ \alpha & \alpha & \alpha+\delta \\ \beta & \beta & \beta+\gamma \end{vmatrix}$.
Since the first and second columns are identical,the value of the determinant is $0$.
Thus,the given lines are coplanar.