⇒Vectors

Correct answer option is (B)

$\begin{array}{ll}& \bar{\mathrm{a}}+2\bar{\mathrm{b}}\text{ is collinear with }\bar{\mathrm{c}}\\ \therefore & \bar{\mathrm{a}}+2\bar{\mathrm{b}}=\mathrm{nc}\text{.... (i)}\\ & \text{ Similarly }\bar{\mathrm{b}}+3\bar{\mathrm{c}}=\mathrm{ma}\text{.... (ii)}\\ & \mathrm{m}\text{ and }\mathrm{n}\text{ are non-zero scalars. }\\ \therefore & \text{ (i) }\Rightarrow \bar{\mathrm{a}}+2\bar{\mathrm{b}}+6\bar{\mathrm{c}}=(\mathrm{n}+6)\bar{\mathrm{c}}\\ & \text{ (ii) }\Rightarrow \bar{\mathrm{a}}+2\bar{\mathrm{b}}+6\bar{\mathrm{c}}=(2\mathrm{m}+1)\bar{\mathrm{a}}\\ & \Rightarrow \mathrm{n}+6=0\text{ and }2\mathrm{m}+1=0\\ & \Rightarrow \mathrm{n}=-6\text{ and }\mathrm{m}=\frac{-1}{2}\\ \therefore & \text{ (i) }\Rightarrow \overrightarrow{a}+2\overrightarrow{b}=-6\bar{\mathrm{c}}\end{array}$

Correct answer option is (A)

Centroid of tetrahedron

$\begin{array}{rl}& \equiv (\frac{-1+3+2-1}{4},\frac{2-2+1-2}{4},\frac{3+1+3+4}{4})\\ & \equiv (\frac{3}{4},\frac{-1}{4},\frac{11}{4})\end{array}$

Correct answer option is (A)

$\begin{array}{rl}\overline{a}+\overline{b}& =(\hat{i}+\hat{j}+\hat{k})+(3\hat{i}-2\hat{j}+5\hat{k})\\ & =4\hat{i}-\hat{j}+6\hat{k}\\ \overline{a}-\overline{b}& =(\hat{i}+\hat{j}+\hat{k})-(3\hat{i}-2\hat{j}+5\hat{k})\\ & =-2\hat{i}+3\hat{j}-4\hat{k}\end{array}$

$\therefore$ Vector perpendicular to $(\overline{a}+\overline{b})$ and $(\overline{a}-\overline{b})$ is

$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& -1& 6\\ -2& 3& -4\end{array}\right|=-14\hat{i}+4\hat{j}+10\hat{k}$

$\therefore$ Required unit vector is

$\frac{-14\hat{\mathrm{i}}+4\hat{\mathrm{j}}+10\hat{\mathrm{k}}}{\sqrt{(-14{)}^2+4^2+(10{)}^2}}=\frac{-14\hat{\mathrm{i}}+4\hat{\mathrm{j}}+10\hat{\mathrm{k}}}{\sqrt{312}}$

Correct answer option is (A)

Note that only for option (A), i.e., for $\alpha =1$ and $\beta =1,|\overrightarrow{c}|=\sqrt{3}$ holds true.

$\therefore$ Option (A) is correct.

Correct answer option is (D)

Let $\overline{r}=a\hat{i}+b\hat{j}+c\hat{k}$

As $\bar{\mathrm{r}}$ is perpendicular to $\bar{\mathrm{q}}$.

$\begin{array}{rl}\therefore & \overline{r}\cdot \overline{q}=0\\ & \Rightarrow a-2b+c=0\text{.... (i)}\end{array}$

Also, $\overline{r}$ is coplanar with vectors $\overline{p}$ and $\overline{q}$

$\begin{array}{rl}& \therefore \left[\begin{array}{ccc}\overline{p}& \overline{q}& \overline{r}\end{array}\right]=0\\ \Rightarrow & \left|\begin{array}{ccc}1& 1& 1\\ 1& -2& 1\\ a& b& c\end{array}\right|=0\\ & \Rightarrow 3a-3c=0\\ & \Rightarrow a-c=0\\ & \Rightarrow a=c\text{..... (ii)}\end{array}$

From (i) and (ii), we get

$\begin{array}{rl}\mathrm{b}& =\mathrm{c}\\ \therefore \bar{\mathrm{r}}& =\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\end{array}$

Now, the magnitude of required vector is $5\sqrt{3}$ units.

$\begin{array}{rl}\therefore \text{ Required vector }& =5\sqrt{3}\times \frac{\bar{\mathrm{r}}}{|\overrightarrow{\mathrm{r}}|}\\ & =5\sqrt{3}\times \frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{3}}=5(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})\end{array}$