(3)

Position vector of $A=2\hat{i}+2\hat{j}+\hat{k}$

Position vector of $B=\hat{i}+2\hat{j}+2\hat{k}$

Position vector of $C=2\hat{i}+\hat{j}+2\hat{k}$

$\Rightarrow$ Coordinates of A, B and C of $△ABC$ are A(2, 2, 1), B(1, 2, 2) and C(2, 1, 2)

$AB=\sqrt{{(1–2)}^2+{(2–2)}^2+{(2–1)}^2}=\sqrt{2} \mathrm{units}$

$BC=\sqrt{{(2–1)}^2+{(1–2)}^2+{(2–2)}^2}=\sqrt{2} \mathrm{units}$

$AC=\sqrt{{(2–2)}^2+{(1–2)}^2+{(2–1)}^2}=\sqrt{2} \mathrm{units}$

$\Rightarrow$ $△ABC$ is an equilateral triangle with side $\sqrt{2}$ units. In an equilateral triangle, orthocenter and centroid will be same.

Centroid, $G(x,y,z)\equiv G(\frac{2+1+2}{3},\frac{2+2+1}{3},\frac{2+2+1}{3})\equiv G(\frac{5}{3},\frac{5}{3},\frac{5}{3})$

Mid-point of AB is $D(\frac{3}{2},2,\frac{3}{2})$

$l_1=\sqrt{{(\frac{5}{3}-\frac{3}{2})}^2+{(\frac{5}{3}–2)}^2+{(\frac{5}{3}–\frac{3}{2})}^2}=\sqrt{\frac{1}{36}+\frac{1}{9}+\frac{1}{36}}=\sqrt{\frac{1}{6}}$

    $l_1=l_2=l_3=\sqrt{\frac{1}{6}}$

$\therefore l_1^2+l_2^2+l_3^2=3\times \frac{1}{6}=\frac{1}{2}$

(3)

In ,$△ABC$

Position vector of $A=2\hat{i}–3\hat{j}+3\hat{k}$

Position vector of $B=2\hat{i}+2\hat{j}+3\hat{k}$

Position vector of $C=–\hat{i}+\hat{j}+3\hat{k}$

Coordinates of the triangle ABC are A(2, –3, 3), B(2, 2, 3) and C(–1, 1, 3).

AD is the angle bisector of $\angle BAC$.

$\Rightarrow BD:DC=AB:AC\Rightarrow D$ is the mid-point of the side BC.

Let the coordinates of D be (x, y, z).

By mid-point formula, $x=\frac{2–1}{2}, y=\frac{2+1}{2}, z=\frac{3+3}{2}$

$\Rightarrow x=\frac{1}{2}, y=\frac{3}{2} \mathrm{and} z=3$

So, $A{D}^2={(2–\frac{1}{2})}^2+{(–3–\frac{3}{2})}^2+{(3–3)}^2=\frac{90}{4}$

$\Rightarrow l^2=A{D}^2=\frac{90}{4}$               $\therefore 2l^2=2\times \frac{90}{4}=45$

(3)

$\left|\overrightarrow{OA}\right|$$=\sqrt{4+4+1}=3$

$\left|\overrightarrow{OB}\right|$$=\sqrt{4+16+16}=6$

$\frac{OA}{OB}=\frac{AC}{CB}=\frac{3}{6}=\frac{1}{2}$

$C\equiv (\frac{4+2}{3},\frac{4+4}{3},\frac{2+4}{3})$

$\therefore C\equiv (2,\frac{8}{3},2)$

$\left|\overrightarrow{OC}\right|$$=\sqrt{4+\frac{64}{9}+4}=\sqrt{\frac{136}{3}}=\frac{2\sqrt{34}}{3}$.

(1)

We have

 $\overrightarrow{a}+5\overrightarrow{b}=\lambda \overrightarrow{c}, \lambda \in R$         ... (i)

$\overrightarrow{b}+6\overrightarrow{c}=\mu \overrightarrow{a}, \mu \in R \Rightarrow \overrightarrow{a}=\frac{\overrightarrow{b}+6\overrightarrow{c}}{\mu }$

From (i), $\frac{\overrightarrow{b}+6\overrightarrow{c}}{\mu }+5\overrightarrow{b}=\lambda \overrightarrow{c} \Rightarrow \overrightarrow{b}(\frac{1}{\mu }+5)=\overrightarrow{c}(\lambda –\frac{6}{\mu })$

Since, $\overrightarrow{b}$ and $\overrightarrow{c}$ are non-collinear.

$\therefore \frac{1}{\mu }=–5 \mathrm{and} \lambda =\frac{6}{\mu } \Rightarrow \mu =\frac{–1}{5} \mathrm{and} \lambda =–30$

$\therefore \mathrm{From} \left(\mathrm{i}\right), \overrightarrow{a}+5\overrightarrow{b}–\lambda \overrightarrow{c}=0$

$\Rightarrow \overrightarrow{a}+5\overrightarrow{b}+30\overrightarrow{c} =0\Rightarrow \alpha =5 \mathrm{and} \beta =30 \therefore \alpha +\beta =35$