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**Hint:**In this question, we are given three non-coplanar vectors and we need to find the value of $ \dfrac{\left( a+2b-c \right)\cdot \left( a-b \right)\times \left( a-b-c \right)}{\left[ abc \right]} $ . For this we will first find the cross product between (a-b) and (a-b-c). Then we will solve the dot product and simplify our answer. We will use the following properties of dot product and cross product:

(I) $ a\times b=-b\times a $ commutative property is not applied in cross product.

(II) $ a\times a=0,b\times b=0,c\times c=0 $ .

(III) $ a\cdot b=b\cdot a $ commutative property is applied in dot product.

(IV) [a,b,c] defines triple product and can be denoted by $ a\cdot \left( b\times c \right),b\cdot \left( c\times a \right),c\cdot \left( a\times b \right) $ .

In a triple product if two of the vectors are same then [a,b,c] =0 i.e. [aab] = 0 or [abb] = 0 and so on.

**Complete step by step answer:**

Here we are given three non-coplanar vectors a, b, c, we need to find the value of $ \dfrac{\left( a+2b-c \right)\cdot \left( a-b \right)\times \left( a-b-c \right)}{\left[ abc \right]}\cdots \cdots \cdots \left( 1 \right) $ .

Since a, b, c are non-coplanar vectors therefore [a,b,c] is not equal to zero. Now let us solve the cross product between (a-b) and (a-b-c) first.

We have $ \begin{align}

& \left( a-b \right)\times \left( a-b-c \right)=a\times \left( a-b-c \right)-b\times \left( a-b-c \right) \\

& \Rightarrow \left( a-b \right)\times \left( a-b-c \right)=a\times a-a\times b-a\times c-b\times a+b\times b+b\times c \\

\end{align} $ .

We know that the cross product between same vectors is equal to zero. Therefore, we can say $ a\times a=0\text{ and }b\times b=0 $ so we have,

$ \left( a-b \right)\times \left( a-b-c \right)=0-a\times b-a\times c-b\times a+0+b\times c $ .

Now we know that in cross product $ a\times b=-b\times a $ so using this we get,

$ \left( a-b \right)\times \left( a-b-c \right)=-a\times b-a\times c+a\times b+b\times c $ .

Cancelling $ a\times b $ with $ -a\times b $ we get,

$ \left( a-b \right)\times \left( a-b-c \right)=b\times c-a\times c $ .

For simplification let us use $ -a\times c=c\times a $ we get,

$ \left( a-b \right)\times \left( a-b-c \right)=b\times c+c\times a $ .

Putting this value in the given expression (1) we get,

$ \dfrac{\left( a+2b-c \right)\cdot \left( b\times c+c\times a \right)}{\left[ abc \right]} $ .

Let us solve the dot product in the numerator we get,

$ \dfrac{a\cdot \left( b\times c \right)+a\cdot \left( c\times a \right)+2b\cdot \left( b\times c \right)+2b\cdot \left( c\times a \right)-c\cdot \left( b\times c \right)-c\cdot \left( c\times a \right)}{\left[ abc \right]} $ .

Now we know that $ a\cdot \left( b\times c \right)=b\cdot \left( c\times a \right)=c\cdot \left( a\times b \right) $ are values of triple product [a,b,c] and this two the vectors in [a,b,c] are same, then [a,b,c] = 0. So, we have $ a\cdot \left( c\times a \right)=\left[ aca \right]=0,b\cdot \left( b\times c \right)=\left[ bbc \right]=0,c\cdot \left( b\times c \right)=\left[ cbc \right]=0\text{ and }c\cdot \left( c\times a \right)=\left[ cca \right]=0 $ we get,

$ \dfrac{a\cdot \left( b\times c \right)+0+0+2b\cdot \left( c\times a \right)-0-0}{\left[ abc \right]} $ .

As discussed earlier $ a\cdot \left( b\times c \right)=b\cdot \left( c\times a \right)=\left[ abc \right] $ so we get,

$ \dfrac{\left[ abc \right]+2\left[ abc \right]}{\left[ abc \right]}\Rightarrow \dfrac{3\left[ abc \right]}{\left[ abc \right]} $ .

Cancelling [a,b,c] from the numerator and the denominator we get, 3 which is our required value.

Hence $ \dfrac{\left( a+2b-c \right)\cdot \left( a-b \right)\times \left( a-b-c \right)}{\left[ abc \right]}=3 $ .

**Note:**

Students should keep in mind the properties of the cross product and the dot product of vectors. Note that, if it is given that a, b, c are non-coplanar vectors because if they were coplanar then [a,b,c] would be zero which makes the given expression not defined. Take care of the signs while solving the given expression.

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