If \( \vec{a}=\hat{i}+\lambda \hat{j}+2 \hat{k} ; \vec{b}=\mu \hat{i}+\hat{j}-\hat{k} \) are orthogonal and \( |\vec{a}|=|\vec{b}| \) then
\( (\lambda, \mu)= \)

Given vectors,
\[
\begin{array}{l}
\vec{a}=\hat{i}+\lambda \hat{j}+2 \hat{k} \rightarrow(1) \\
\vec{b}=\mu \hat{i}+\hat{j}-\hat{k} \rightarrow(2)
\end{array}
\]
For vectors to be orthogonal, we have
\[
\begin{array}{l}
\vec{a} \cdot \vec{b}=0 \\
\Rightarrow(\hat{i}+\lambda \hat{j}+2 \hat{k}) \cdot(\mu \hat{i}+\hat{j}-\hat{k})=0 \\
\Rightarrow \mu+\lambda-2=0 \\
\Rightarrow \lambda+\mu=2 \rightarrow(3) \\
\text { and }|\vec{a}|=|\vec{b}| \\
\Rightarrow \sqrt{1+\lambda^{2}+4}=\sqrt{\mu^{2}+1+1} \\
\Rightarrow 1+\lambda^{2}+4=\mu^{2}+1+1 \rightarrow(4)
\end{array}
\]
From Eqs. (3) and (4), we get
\[
\begin{array}{l}
\lambda^{2}+3=\mu^{2}=(2 \lambda)^{2} \\
\Rightarrow \lambda^{2}+3=4+\lambda^{2}-4 \lambda \\
\text { So, } \lambda=\frac{1}{4} \Rightarrow \mu=2-\frac{1}{4}=\frac{7}{4}
\end{array}
\]
\[
\text { Hence }(\lambda, \mu)=\left(\frac{1}{4}, \frac{7}{4}\right)
\]