ABCD is a trapezium in which AB and CD are parallel sides. If \[l\left( AB \right)\text{ }=\text{ }3l\left( CD \right)\] and $\overrightarrow{DC}={2\widehat{i}-5\widehat{k}}$. Then vector $\overrightarrow{AB}$ is
(a) ${\dfrac{1}{\sqrt{29}}\left( 2\widehat{i}-5\widehat{k} \right)}$
(b) ${\dfrac{\sqrt{29}}{3}\left( 5\widehat{i}-2\widehat{k} \right)}$
(c) ${-6\widehat{i}+15\widehat{k}}$
(d) a or b

Hint: When one vector is parallel to another vector then their direction is the same but magnitude is different. So, by using the above relation we can establish a distinct relation leading to our result.

Complete step-by-step answer:
We are provided with a vector $\overrightarrow{DC}=2\widehat{i}-5\widehat{k}$ as one of the sides of a trapezium ABCD. The other side which is given as parallel to $\overrightarrow{DC}$.
Using $\overrightarrow{DC}=2\widehat{i}-5\widehat{k}$ we can calculate $\overrightarrow{CD}$ as shown below:
An important property which must be observed in the question is that the direction of the $\overrightarrow{CD}\text{ and }\overrightarrow{DC}$ are opposite. So, in that case multiply $\overrightarrow{CD}$ with subtraction sign (-) to change the direction of the vector $\overrightarrow{CD}$.
Therefore, the vector \[\overrightarrow{CD}\] can be expressed as $\overrightarrow{CD}=-2\widehat{i}+5\widehat{k}$.
AB is parallel to CD as given in the question.
When a vector is parallel to another vector then their mathematical relationship can be stated as:
$\overrightarrow{PQ}=\lambda \overrightarrow{QR}$
So, using this relation for our problem we get,
$\overrightarrow{AB}=\lambda \overrightarrow{CD}$
Putting the value of $\overrightarrow{CD}$ as obtained previously in the above equation we get,
\[\begin{align}
  & \overrightarrow{AB}=\lambda (-2\widehat{i}+5\widehat{k}) \\
 & \overrightarrow{AB}=-2\lambda \widehat{i}+5\lambda \widehat{k}. \\
\end{align}\]
Using condition \[l\left( AB \right)\text{ }=\text{ }3l\left( CD \right)\] stated in the question we can proceed to find $\overrightarrow{AB}$ in terms of $\lambda \overrightarrow{CD}$.
Now, $l(AB)=3l(CD)$ as mentioned above in the question.
For a vector \[\overrightarrow{A}=a\widehat{i}+b\widehat{j}+c\widehat{k}\] the value of length is the same as the value of modulus.
So, modulus of vector \[\overrightarrow{A}=a\widehat{i}+b\widehat{j}+c\widehat{k}\] can be expressed as:
\[\left| \overrightarrow{A} \right|=\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\]
So, putting the values of both $\overrightarrow{AB}\text{ and }\overrightarrow{CD}$ in the modulus definition we get,
$\begin{align}
  & \sqrt{{{(-2\lambda )}^{2}}+{{(5\lambda )}^{2}}}=3\sqrt{{{(-2)}^{2}}+{{(5)}^{2}}} \\
 & \sqrt{4{{\lambda }^{2}}+25{{\lambda }^{2}}}=3\sqrt{4+25} \\
 & \sqrt{29{{\lambda }^{2}}}=3\sqrt{29} \\
 & \sqrt{29}\lambda =3\sqrt{29} \\
\end{align}$
On comparing both sides of the equation we obtained $\lambda =3$.
After finding the value of $\lambda $ we replace it in the expression of $\overrightarrow{AB}$ to find the actual numeric value of $\overrightarrow{AB}$ ,
\[\begin{align}
  & \overrightarrow{AB}=-2\times 3\widehat{i}+5\times 3\widehat{k} \\
 & \overrightarrow{AB}=-6\widehat{i}+15\widehat{k}. \\
\end{align}\]
$\therefore $The value of \[\overrightarrow{AB}\text{ is }-6\widehat{i}+15\widehat{k}\].
Therefore, option (c) is correct.

Note: We must convert $\overrightarrow{CD}\text{ and }\overrightarrow{DC}$ vectors as most of the student will attempt the question as $\overrightarrow{DC}$ which invokes the basic error of sign convention of vectors. The key step is to express the vectors in the parallel form correctly.Students should remember that when a vector is parallel to another vector then their mathematical relationship can be stated as: $\overrightarrow{PQ}=\lambda \overrightarrow{QR}$