If $\overrightarrow{a}$ and $\overrightarrow{b}$ are two vectors, then \[\left( 2\overrightarrow{a}+3\overrightarrow{b} \right)\times \left( 5\overrightarrow{a}+7\overrightarrow{b} \right)+\left( \overrightarrow{a}\times \overrightarrow{b} \right)\] is equal to?
(a) $0$
(b) $1$
(c) \[\overrightarrow{a}\times \overrightarrow{b}\]
(d) \[\overrightarrow{b}\times \overrightarrow{a}\]

Hint: Assume the given expression as E. Take the cross product \[\left( 2\overrightarrow{a}+3\overrightarrow{b} \right)\times \left( 5\overrightarrow{a}+7\overrightarrow{b} \right)\] and simplify it using the relations \[\overrightarrow{a}\times \overrightarrow{a}=0\] and \[\overrightarrow{b}\times \overrightarrow{b}=0\]. Further, use the relation \[\overrightarrow{b}\times \overrightarrow{a}=-\overrightarrow{a}\times \overrightarrow{b}\] by considering the fact that the rotation of vectors in the two cases are in opposite direction and the vector obtained by the vector product is perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{b}$. Add the term \[\left( \overrightarrow{a}\times \overrightarrow{b} \right)\] present at the end to get the answer.

Complete step-by-step solution:
Here we have been provided with two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$, we are asked to find the value of the expression \[\left( 2\overrightarrow{a}+3\overrightarrow{b} \right)\times \left( 5\overrightarrow{a}+7\overrightarrow{b} \right)+\left( \overrightarrow{a}\times \overrightarrow{b} \right)\]. Let us assume the expression as E, so we have,
\[\Rightarrow E=\left( 2\overrightarrow{a}+3\overrightarrow{b} \right)\times \left( 5\overrightarrow{a}+7\overrightarrow{b} \right)+\left( \overrightarrow{a}\times \overrightarrow{b} \right)\]
Now, first considering the cross product \[\left( 2\overrightarrow{a}+3\overrightarrow{b} \right)\times \left( 5\overrightarrow{a}+7\overrightarrow{b} \right)\] we have,
\[\Rightarrow E=\left\{ 10\left( \overrightarrow{a}\times \overrightarrow{a} \right)+14\left( \overrightarrow{a}\times \overrightarrow{b} \right)+15\left( \overrightarrow{b}\times \overrightarrow{a} \right)+21\left( \overrightarrow{b}\times \overrightarrow{b} \right) \right\}+\left( \overrightarrow{a}\times \overrightarrow{b} \right)\]
We know that \[\left( \overrightarrow{a}\times \overrightarrow{b} \right)=ab\sin \theta \left( {\hat{n}} \right)\], where a and b are the magnitudes of $\overrightarrow{a}$ and $\overrightarrow{b}$ respectively, $\theta $ is the angle between the two vectors and \[\left( {\hat{n}} \right)\] is a unit vector perpendicular to the planes of both $\overrightarrow{a}$ and $\overrightarrow{b}$. Since the angle between two equal vectors is ${{0}^{\circ }}$ and we have $\sin {{0}^{\circ }}=0$, therefore \[\left( \overrightarrow{a}\times \overrightarrow{a} \right)=0\] and \[\left( \overrightarrow{b}\times \overrightarrow{b} \right)=0\].
\[\Rightarrow E=\left\{ 14\left( \overrightarrow{a}\times \overrightarrow{b} \right)+15\left( \overrightarrow{b}\times \overrightarrow{a} \right) \right\}+\left( \overrightarrow{a}\times \overrightarrow{b} \right)\]
In the cross product \[\left( \overrightarrow{a}\times \overrightarrow{b} \right)\] we rotate $\overrightarrow{a}$ towards $\overrightarrow{b}$ and in cross product \[\left( \overrightarrow{b}\times \overrightarrow{a} \right)\] we rotate $\overrightarrow{b}$ towards $\overrightarrow{a}$, so in the two cases the vectors obtained are perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{b}$ but they are opposite in direction. If the unit vector along \[\left( \overrightarrow{a}\times \overrightarrow{b} \right)\] is $\hat{n}$ then the unit vector along \[\left( \overrightarrow{b}\times \overrightarrow{a} \right)\] will be $-\hat{n}$. Therefore, we have the relation \[\overrightarrow{b}\times \overrightarrow{a}=-\overrightarrow{a}\times \overrightarrow{b}\], so we get,
\[\begin{align}
  & \Rightarrow E=\left\{ 14\left( \overrightarrow{a}\times \overrightarrow{b} \right)-15\left( \overrightarrow{a}\times \overrightarrow{b} \right) \right\}+\left( \overrightarrow{a}\times \overrightarrow{b} \right) \\
 & \Rightarrow E=-\left( \overrightarrow{a}\times \overrightarrow{b} \right)+\left( \overrightarrow{a}\times \overrightarrow{b} \right) \\
 & \therefore E=0 \\
\end{align}\]
Hence, option (a) is the correct answer.

Note: Note that the cross product results in a vector quantity and that is why it is also known as the vector product, while the dot product results in a scalar quantity and hence it is called the scalar product. In case of dot product we have the relation \[\overrightarrow{a}.\overrightarrow{b}=ab\cos \theta \] and it is equal to 0 when the two vectors are perpendicular to each other.