If \(\overline{\mathrm{p}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\) and \(\overline{\mathrm{q}}=\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\). Then a vector of magnitude \(5 \sqrt{3}\) units perpendicular to the vector \(\overline{\mathrm{q}}\) and coplanar with \(\overline{\mathrm{p}}\) and \(\overline{\mathrm{q}}\) is

Let \(\bar{r}=a \hat{i}+b \hat{j}+c \hat{k}\)
As \(\overline{\mathrm{r}}\) is perpendicular to \(\overline{\mathrm{q}}\).
\(\begin{array}{ll}
\therefore \quad \bar{r} \cdot \bar{q}=0 \\
\quad \Rightarrow a-2 b+c=0 ...(i)
\end{array}\)
Also, \(\bar{r}\) is coplanar with vectors \(\bar{p}\) and \(\bar{q}\)
\(\begin{aligned}
\therefore \quad & {\left[\begin{array}{ccc}
\bar{p} & \overline{\mathrm{q}} & -\mathrm{r}
\end{array}\right]=0 } \\
& \Rightarrow\left|\begin{array}{ccc}
1 & 1 & 1 \\
1 & -2 & 1 \\
\mathrm{a} & \mathrm{b} & \mathrm{c}
\end{array}\right|=0 \\
& \Rightarrow 3 \mathrm{a}-3 \mathrm{c}=0 \\
& \Rightarrow \mathrm{a}-\mathrm{c}=0 \\
& \Rightarrow \mathrm{a}=\mathrm{c}...(ii)
\end{aligned}\)
From (i) and (ii), we get
\(\mathrm{b}=\mathrm{c}\)
\(\therefore -\overline{\mathrm{r}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}^{\mathrm{s}}\)
Now, the magnitude of required vector is \(5 \sqrt{3}\) units.
\(\text {Required vector } =5 \sqrt{3} \times \frac{r}{\sqrt{r}} \)
\( =5 \sqrt{3} \times \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{3}}=5(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})\)