If $\overrightarrow{a}+2\overrightarrow{b}+3\overrightarrow{c}=0,$ then $\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}=k\overrightarrow{a}\times \overrightarrow{b},$ Where k is equal to ?
A.0
B.1
C.2
D.3

Hint: We are required to find the value of k in the above equation given by $\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}=k\overrightarrow{a}\times \overrightarrow{b},$ provided the equation $\overrightarrow{a}+2\overrightarrow{b}+3\overrightarrow{c}=0,$ We are dealing with vectors in this question and use the concept of cross product here. We multiply the first equation $\overrightarrow{a}+2\overrightarrow{b}+3\overrightarrow{c}=0$ by vectors of a, b and c and get 3 different equations with 3 different relations. We then substitute them in the other equation and find the value of k.

Complete step by step solution:
To solve this question, consider the relation $\overrightarrow{a}+2\overrightarrow{b}+3\overrightarrow{c}=0.$ We take the cross product of the above equation with vector a.
$\Rightarrow \overrightarrow{a}\times \overrightarrow{a}+2\overrightarrow{b}\times \overrightarrow{a}+3\overrightarrow{c}\times \overrightarrow{a}=0\times \overrightarrow{a}$
We know that the cross product of any vector with itself is zero, that is, $\overrightarrow{a}\times \overrightarrow{a}=0.$ We also know that any vector multiplied with 0 gives us 0. Using this,
$\Rightarrow 0+2\overrightarrow{b}\times \overrightarrow{a}+3\overrightarrow{c}\times \overrightarrow{a}=0$
We also know that if the order of the cross-product changes, the sign of it changes too, $\overrightarrow{a}\times \overrightarrow{b}=-\overrightarrow{b}\times \overrightarrow{a}.$ Using this for the $2\overrightarrow{b}\times \overrightarrow{a}$ term,
$\Rightarrow 0-2\overrightarrow{a}\times \overrightarrow{b}+3\overrightarrow{c}\times \overrightarrow{a}=0$
Rearranging,
$\Rightarrow 2\overrightarrow{a}\times \overrightarrow{b}=3\overrightarrow{c}\times \overrightarrow{a}\ldots \ldots \left( 1 \right)$
Repeating the same steps again but this time, we take a cross product with vector b.
$\Rightarrow \overrightarrow{a}\times \overrightarrow{b}+2\overrightarrow{b}\times \overrightarrow{b}+3\overrightarrow{c}\times \overrightarrow{b}=0\times \overrightarrow{b}$
Second term and the term on the right-hand side of the equation become 0. The third term on the left-hand side is rearranged to form a cross product $\overrightarrow{b}\times \overrightarrow{c}.$
$\Rightarrow \overrightarrow{a}\times \overrightarrow{b}+0-3\overrightarrow{b}\times \overrightarrow{c}=0$
Rearranging,
$\Rightarrow \overrightarrow{a}\times \overrightarrow{b}=3\overrightarrow{b}\times \overrightarrow{c}\ldots \ldots \left( 2 \right)$
Now again, we repeat the steps but this time we use the cross product with vector c.
$\Rightarrow \overrightarrow{a}\times \overrightarrow{c}+2\overrightarrow{b}\times \overrightarrow{c}+3\overrightarrow{c}\times \overrightarrow{c}=0\times \overrightarrow{c}$
Third term and the term on the right-hand side of the equation become 0. The first term on the left-hand side is rearranged to form a cross product $\overrightarrow{c}\times \overrightarrow{a}.$
$\Rightarrow -\overrightarrow{c}\times \overrightarrow{a}+2\overrightarrow{b}\times \overrightarrow{c}+0=0$
Rearranging,
$\Rightarrow \overrightarrow{c}\times \overrightarrow{a}=2\overrightarrow{b}\times \overrightarrow{c}\ldots \ldots \left( 3 \right)$
Now consider the equation given in question,
$\Rightarrow \overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}=k\overrightarrow{a}\times \overrightarrow{b}$
Substituting for the first and third terms from equations 2 and 3,
$\Rightarrow 3\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{b}\times \overrightarrow{c}+2\overrightarrow{b}\times \overrightarrow{c}=k\overrightarrow{a}\times \overrightarrow{b}$
Adding the terms on the left-hand side,
$\Rightarrow 6\overrightarrow{b}\times \overrightarrow{c}=k\overrightarrow{a}\times \overrightarrow{b}$
Using equation 2 to substitute in the above equation,
$\Rightarrow 2.3\overrightarrow{b}\times \overrightarrow{c}=k\overrightarrow{a}\times \overrightarrow{b}$
$\Rightarrow 2\overrightarrow{a}\times \overrightarrow{b}=k\overrightarrow{a}\times \overrightarrow{b}$
Comparing both sides of the equation, we get $k=2.$

So, the correct answer is “Option C”.

Note: It is important to know the basics of vectors, cross-product of vectors and dot product of vectors in order to solve such questions. We need to note that the cross product of two vectors is usually considered in a three-dimensional space since the cross product of two perpendicular vectors yield another vector perpendicular to both the original vectors.