Answer:
Option C
Explanation:
Given , $a=\hat{i}+\hat{j}-2\hat{k},b=2\hat{i}-\hat{j}+\hat{k}$ and $c=3\hat{i}-\hat{k}$ and c=ma+nb
$\therefore$ $3\hat{i}-\hat{k}=m(\hat{i}+\hat{j}-2\hat{k})+n(2\hat{i}-\hat{j}+\hat{k})$
$\Rightarrow$ $3\hat{i}-\hat{k}=(m+2n)\hat{i}+(m-n)\hat{j}+(-2m+n)\hat{k}$
On equating the coefficient of $\hat{i}$, $\hat{j}$ and $\hat{k}$ , respectively
we get
3=m+2n,0=m-n
and -1=-2m+n
$\Rightarrow$ 3=n+2n
$\Rightarrow$ n=1
$\Rightarrow$ m=1 and n=1
$\Rightarrow$ m+n=1+1=2
