The vector of magnitude 6 units and perpendicular to vectors \(2 \hat{i}+\hat{j}-3 \hat{k}\) and \(\hat{i}-2 \hat{j}+\hat{k}\) is

Let the required vector be \(\overline{\mathrm{r}}=x \hat{\mathrm{i}}+y \hat{\mathrm{j}}+z \hat{\mathrm{k}}\).
Then, \(|\hat{r}|=6\)
\(\Rightarrow x^2+y^2+z^2=36\)
Now, \(\bar{r}\) is perpendicular to vectors \(\bar{a}=2 \hat{i}+\hat{j}-3 \hat{k}\) and \(\bar{b}=\hat{i}-2 \hat{j}+\hat{k}\).
\(\begin{array}{ll}
\therefore & \overline{\mathrm{a}} \times \overline{\mathrm{b}}=\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
2 & 1 & -3 \\
1 & -2 & 1
\end{array}\right|=-5(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}) \\
\therefore & x=y=\mathrm{z} \\
\therefore & \text { Let } x=y=\mathrm{z}=\lambda
\end{array}\)
Let \(x=y=z=\lambda\).
From (i), we get
\(3 \lambda^2=36 \)
\( \therefore \lambda=2 \sqrt{3}\)
\(\therefore\) Required vector is \(2 \sqrt{3}(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})\)