Are lines $\dfrac{x-a+d}{\alpha -\delta }=\dfrac{y-a}{\alpha }=\dfrac{z-a-d}{\alpha +\delta }$ and $\dfrac{x-b+c}{\beta -\gamma }=\dfrac{y-b}{\beta }=\dfrac{z-b-c}{\beta +\gamma }$ coplanar? Justify!

Hint : To justify the given question, we will use the method of determinant. We will use the condition of coplanar of two lines, that is if the lines are coplanar then the determinant of its elements will be equal to zero. So, we will use the reverse method. We will check if the determinant of its elements is zero or not. If it is zero, the lines will be coplanar.

Complete step-by-step solution:
As we know that two lines $\dfrac{x-{{x}_{1}}}{{{l}_{1}}}=\dfrac{y-{{y}_{1}}}{{{m}_{1}}}=\dfrac{z-{{z}_{1}}}{{{n}_{1}}}$ and $\dfrac{x-{{x}_{2}}}{{{l}_{2}}}=\dfrac{y-{{y}_{2}}}{{{m}_{2}}}=\dfrac{z-{{z}_{2}}}{{{n}_{2}}}$ are coplanar if $\left| \begin{matrix}
   \left( {{x}_{2}}-{{x}_{1}} \right) & \left( {{y}_{2}}-{{y}_{1}} \right) & \left( {{z}_{2}}-{{z}_{1}} \right) \\
   {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \\
   {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \\
\end{matrix} \right|=0$.
So, to justify that the given lines are coplanar, we will use this property.
Since, we will compare the given line to the assumed lines to get the relative values.
Since, the given lines are lines $\dfrac{x-a+d}{\alpha -\delta }=\dfrac{y-a}{\alpha }=\dfrac{z-a-d}{\alpha +\delta }$ and $\dfrac{x-b+c}{\beta -\gamma }=\dfrac{y-b}{\beta }=\dfrac{z-b-c}{\beta +\gamma }$.
We can simplify it as:
$\Rightarrow \dfrac{x-\left( a-d \right)}{\alpha -\delta }=\dfrac{y-a}{\alpha }=\dfrac{z-\left( a+d \right)}{\alpha +\delta }$ and $\dfrac{x-\left( b-c \right)}{\beta -\gamma }=\dfrac{y-b}{\beta }=\dfrac{z-\left( b+c \right)}{\beta +\gamma }$
After comparing with $\dfrac{x-{{x}_{1}}}{{{l}_{1}}}=\dfrac{y-{{y}_{1}}}{{{m}_{1}}}=\dfrac{z-{{z}_{1}}}{{{n}_{1}}}$ and $\dfrac{x-{{x}_{2}}}{{{l}_{2}}}=\dfrac{y-{{y}_{2}}}{{{m}_{2}}}=\dfrac{z-{{z}_{2}}}{{{n}_{2}}}$, we will have:
\[\begin{align}
  & {{x}_{1}}=a-d \\
 & {{x}_{2}}=b-c \\
 & {{y}_{1}}=a \\
 & {{y}_{2}}=b \\
 & {{z}_{1}}=a+d \\
 & {{z}_{2}}=b+c \\
 & {{l}_{1}}=\alpha -\delta \\
 & {{l}_{2}}=\beta -\gamma \\
 & {{m}_{1}}=\alpha \\
 & {{m}_{2}}=\beta \\
 & {{n}_{1}}=\alpha +\delta \\
 & {{n}_{2}}=\beta +\gamma \\
\end{align}\]
Now, we will use the coplanar condition as:
$\Rightarrow \left| \begin{matrix}
   \left( {{x}_{2}}-{{x}_{1}} \right) & \left( {{y}_{2}}-{{y}_{1}} \right) & \left( {{z}_{2}}-{{z}_{1}} \right) \\
   {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \\
   {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \\
\end{matrix} \right|$
Now, we will substitute the corresponding values in the above step as:
$\Rightarrow \left| \begin{matrix}
   \left( b-c-\left( a-d \right) \right) & \left( b-a \right) & \left( b+c-\left( a+d \right) \right) \\
   \alpha -\delta & \alpha & \alpha +\delta \\
   \beta -\gamma & \beta & \beta +\gamma \\
\end{matrix} \right|$
Here, we will open the bracket as:
$\Rightarrow \left| \begin{matrix}
   \left( b-c-a+d \right) & \left( b-a \right) & \left( b+c-a-d \right) \\
   \alpha -\delta & \alpha & \alpha +\delta \\
   \beta -\gamma & \beta & \beta +\gamma \\
\end{matrix} \right|$
Now, we will use the rule of determinants to simplify it. Here, we will change column first by adding first column and third column, ${{C}_{1}}\to {{C}_{1}}+{{C}_{3}}$, as:
$\Rightarrow \left| \begin{matrix}
   \left( b-c-a+d+b+c-a-d \right) & \left( b-a \right) & \left( b+c-a-d \right) \\
   \alpha -\delta +\alpha +\delta & \alpha & \alpha +\delta \\
   \beta -\gamma +\beta +\gamma & \beta & \beta +\gamma \\
\end{matrix} \right|$
Now, we will cancel out the equal like terms as:
\[\Rightarrow \left| \begin{matrix}
   2\left( b-a \right) & \left( b-a \right) & \left( b+c-a-d \right) \\
   2\alpha & \alpha & \alpha +\delta \\
   2\beta & \beta & \beta +\gamma \\
\end{matrix} \right|\]
Here, we will divide by $2$in column first, ${{C}_{1}}\to \dfrac{{{C}_{1}}}{2}$ as:
\[\Rightarrow \left| \begin{matrix}
   \left( b-a \right) & \left( b-a \right) & \left( b+c-a-d \right) \\
   \alpha & \alpha & \alpha +\delta \\
   \beta & \beta & \beta +\gamma \\
\end{matrix} \right|\]
Since, two columns are the same. So the determinant will be zero by the property of the determinant.
$\Rightarrow 0$
Hence, the given lines are coplanar.

Note:Method of solution of determinant of a matrix $3\times 3$ as:
Let’s the determinant of a matrix is writes as:
$\Rightarrow \left| \begin{matrix}
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{a}_{4}} & {{a}_{5}} & {{a}_{6}} \\
   {{a}_{7}} & {{a}_{8}} & {{a}_{9}} \\
\end{matrix} \right|$
Now, the determinant of matrix can be obtained as:
$\Rightarrow \left| \begin{matrix}
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{a}_{4}} & {{a}_{5}} & {{a}_{6}} \\
   {{a}_{7}} & {{a}_{8}} & {{a}_{9}} \\
\end{matrix} \right|={{a}_{1}}\left( {{a}_{5}}\times {{a}_{9}}-{{a}_{8}}\times {{a}_{6}} \right)-{{a}_{2}}\left( {{a}_{4}}\times {{a}_{9}}-{{a}_{7}}\times {{a}_{6}} \right)+{{a}_{3}}\left( {{a}_{4}}\times {{a}_{8}}-{{a}_{7}}\times {{a}_{5}} \right)$