please help me, a=2i+j+k,b=i+2j+kand c=i+j+2k be three vecters. A vector in the plane of b and c whose projection on a is underroot 2/3 will be,

1. 2i+2j-3k
2. 2i+3j+k
3. -2i-j+5k
4. 2i+j+5k

a vector in the plane of b and c is 
$$\begin{gathered} d=\alpha b+\beta c \\ =\alpha .(i+2j+k)+\beta .(i+j+2k) \\ =(\alpha +\beta ).i+(2\alpha +\beta ).j+(\alpha +2\beta )k \end{gathered}$$
projection of d on a is $\frac{d.a}{\left|a\right|}$
$$\begin{gathered} \frac{d.a}{\left|a\right|}=\sqrt{\frac{2}{3}} \\ \frac{\left(\right(\alpha +\beta ).i+(2\alpha +\beta ).j+(\alpha +2\beta \left)k\right).(2i+j+k)}{\left|2i+j+k\right|}=\sqrt{\frac{2}{3}} \\ \frac{2.(\alpha +\beta )+(2\alpha +\beta )+(\alpha +2\beta )}{\sqrt{4+1+1}}=\sqrt{\frac{2}{3}} \\ \frac{5\alpha +5\beta }{\sqrt{6}}=\sqrt{\frac{2}{3}} \\ 5.(\alpha +\beta )=2 \end{gathered}$$
$\Rightarrow \alpha +\beta =\frac{2}{5} ..............\left(1\right)$
then by substituting the different values of $\alpha and \beta$ , we can find the vector d.
let $\alpha =0 and \beta =\frac{2}{5}$
$d=\frac{2}{5}i+\frac{2}{5}j+\frac{4}{5}k$

Dear Kaushik , please again go through the question and do get back to us.
as from the given condition the i component of vector d is 2/5, which does not match with given options.

hope this helps you