⇒Vectors

Correct answer option is (D)

$\bar{\mathrm{a}}$ is $\perp$ er to $\bar{\mathrm{b}}+\bar{\mathrm{c}}\Rightarrow \bar{\mathrm{a}}\cdot (\bar{\mathrm{b}}+\bar{\mathrm{c}})=0\Rightarrow \bar{\mathrm{a}}\cdot \bar{\mathrm{b}}+\bar{\mathrm{a}}\cdot \bar{\mathrm{c}}=0$ ..... (1)

$\bar{\mathrm{b}}$ is $\perp$ er to $(\bar{\mathrm{c}}+\bar{\mathrm{a}})\Rightarrow \bar{\mathrm{b}}\cdot (\bar{\mathrm{c}}+\bar{\mathrm{a}})=0\Rightarrow \bar{\mathrm{b}}\cdot \bar{\mathrm{c}}+\bar{\mathrm{b}}\cdot \bar{\mathrm{a}}=0$ ..... (2)

$\bar{\mathrm{c}}$ is $\perp$ er to $(\bar{\mathrm{a}}+\bar{\mathrm{b}})\Rightarrow \bar{\mathrm{c}}\cdot (\bar{\mathrm{a}}+\bar{\mathrm{b}})=0\Rightarrow \bar{\mathrm{c}}\cdot \bar{\mathrm{a}}+\bar{\mathrm{c}}\cdot \bar{\mathrm{b}}=0$ ...... (3)

From (1), (2) and (3), we get

$2(\bar{\mathrm{a}}\cdot \bar{\mathrm{b}}+\bar{\mathrm{b}}\cdot \bar{\mathrm{c}}+\bar{\mathrm{c}}\cdot \bar{\mathrm{a}})=0$ ...... (4)

Now $|\bar{\mathrm{a}}+\bar{\mathrm{b}}+\bar{\mathrm{c}}{|}^2=|\bar{\mathrm{a}}{|}^2+|\bar{\mathrm{b}}{|}^2+|\mathrm{c}{|}^2+2(\bar{\mathrm{a}}\cdot \bar{\mathrm{b}}+\bar{\mathrm{b}}\cdot \bar{\mathrm{c}}+\bar{\mathrm{c}}\cdot \bar{\mathrm{a}})$

$\therefore |\bar{\mathrm{a}}+\bar{\mathrm{b}}+\bar{\mathrm{c}}{|}^2=(2{)}^2+(3{)}^2+(4{)}^2+0$ ..... [From (4) and data given]

$=4+9+16=29$

$\therefore |\bar{\mathrm{a}}+\bar{\mathrm{b}}+\bar{\mathrm{c}}|=\sqrt{29}$