Let a=2 i- j+ k, b= i+2 j- k and = i+ j-2 k be three vectors. A vector of the type b+λ for some scalar λ, whose projection on a is of magnitude √ 2 3…

Let $\vec{d}=\vec{b}+\lambda \vec{c}$d=b+λc $$\begin{aligned} & \therefore \vec{d}=\hat{i}+2 \hat{j}-\hat{k}+\lambda(\hat{i}+\hat{j}-2 \hat{k}) \\ & =(1+\lambda) \hat{i}+(2+\lambda) \hat{j}-(1+2 \lambda) \hat{k} \end{aligned}$$​∴d=i^+2j^​−k^+λ(i^+j^​−2k^)=(1+λ)i^+(2+λ)j^​−(1+2λ)k^​ If $\theta$θ be the angle between $\vec{d}$d and $\vec{a}$a, then projection of $\vec{d}$d or $(\vec{b}+\lambda \vec{c})$(b+λc) on $\vec{a}$a $$\begin{aligned} & =|\vec{d}| \cos \theta \\ & =|\vec{d}|\left(\frac{\vec{d} \cdot \vec{a}}{|\vec{d}||\vec{a}|}\right)=\frac{\vec{d} \cdot \vec{a}}{|\vec{a}|} \\ & =\frac{2(\lambda+1)-(\lambda+2)-(2 \lambda+1)}{\sqrt{4+1+1}} \\ & =\frac{-\lambda-1}{\sqrt{6}} \end{aligned}$$​=∣d∣cosθ=∣d∣(∣d∣∣a∣d⋅a​)=∣a∣d⋅a​=4+1+1​2(λ+1)−(λ+2)−(2λ+1)​=6​−λ−1​​ But projection of $\vec{d}$d on $\vec{a}=\sqrt{\frac{2}{3}}$a=32​​ $$\therefore-\frac{\lambda+1}{\sqrt{6}}=\sqrt{\frac{2}{3}} \Rightarrow \frac{\lambda^2+2 \lambda+1}{6}=\frac{2}{3}$$∴−6​λ+1​=32​​⇒6λ2+2λ+1​=32​ $$\begin{aligned} & \Rightarrow \lambda^2+2 \lambda-3=0 \Rightarrow \lambda^2+3 \lambda-\lambda-3=0 \\ & \Rightarrow \lambda(\lambda+3)-1(\lambda+3)=0, \Rightarrow \lambda=1,-3 \end{aligned}$$​⇒λ2+2λ−3=0⇒λ2+3λ−λ−3=0⇒λ(λ+3)−1(λ+3)=0,⇒λ=1,−3​ when $\lambda=1$λ=1, then $\vec{b}+\lambda \vec{c}=2 \hat{i}+3 \hat{j}-3 \hat{k}$b+λc=2i^+3j^​−3k^ when $\lambda=-3$λ=−3, then $\vec{b}+\lambda \vec{c}=-2 \hat{i}-\hat{j}+5 \hat{k}$b+λc=−2i^−j^​+5k^