# If three vectors are a(1 ,1,1) ,b (4,-2,3), c(1,-2,1) then find the vector of magnitude 6 unit which is parallel to the vector 2a-b+3c.

$\begin{array}{l}\text{Given that,}\\ \vec{a}=\left(1,1,1\right),\ \vec{b}=\left(4,-2,3\right)\ \&\ \vec{c}=\left(1,-2,1\right)\\ \vec{r}=2\vec{a}-\vec{b}+3\vec{c\ }\\ \vec{r}=2\left(i+j+k\right)-\left(4i-2j+3k\right)+3\left(i-2j+k\right)=i+2k\\ \text{Unit vector is }\left(\vec{r}\right)=\frac{i+2k}{\sqrt{5}}\\ \text{Let}\ \vec{A},\ \text{is parallel to }\vec{r}\ \text{and magnitude is}\ 6.\\ \vec{A}=6\left(\frac{i+2k}{\sqrt{5}}\right)\end{array}$

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