(4)

Line through PQ is given by

$\frac{x–1}{4}=\frac{y+2}{–2}=\frac{z–3}{4}=\lambda$

Any point R on PQ will be $R(4\lambda +1,–2\lambda –2,4\lambda +3)$

Since, PR = 9 units $\Rightarrow {\left(PR\right)}^2=81$

$\Rightarrow {(4\lambda +1–1)}^2+{(–2\lambda –2+2)}^2+{(4\lambda +3–3)}^2=81$

$\Rightarrow 16{\lambda }^2+4{\lambda }^2+16{\lambda }^2=81$

$\Rightarrow 36{\lambda }^2=81 \Rightarrow {\lambda }^2=\frac{9}{4} \Rightarrow \lambda =\pm \frac{3}{2}$

$\therefore$  R can be (7, –5, 9) or (–5, 1, –3)

Distance from origin of the points be $\sqrt{49+25+81}$ and $\sqrt{25+1+9}$i.e., $\sqrt{155},\sqrt{35}$

$\therefore$  Distance of (7, –5, 9) is farthest from origin

$\Rightarrow$$(\alpha ,\beta ,\gamma )\equiv (7, –5, 9)$

Hence ${\alpha }^2+{\beta }^2+{\gamma }^2=7^2+{(–5)}^2+9^2=155$.

(4)

$L_1 : \frac{x–2}{1}=\frac{y–4}{5}=\frac{z–2}{1}=\lambda \left(\mathrm{say}\right)$

$L_2 : \frac{x–3}{2}=\frac{y–2}{3}=\frac{z–3}{2}=\mu \left(\mathrm{say}\right)$

For point of intersection P,

$(\lambda +2,5\lambda +4,\lambda +2)=(2\mu +3,3\mu +2,2\mu +3)$

$\Rightarrow \lambda +2=2\mu +3,5\lambda +4=3\mu +2,\lambda +2=2\mu +3$

$\Rightarrow \lambda =–1,\mu =–1$

$\therefore$  Point P is (1, –1, 1)

Distance of point P from $L_3$ : 4x = 2y = z

$L_3 : \frac{x}{{\displaystyle \frac{1}{4}}}=\frac{y}{{\displaystyle \frac{1}{2}}}=\frac{z}{1}$

Any point on $L_3$ be $R(\frac{\alpha }{4},\frac{\alpha }{2},\alpha )$

D.r.'s of $PR : (\frac{\alpha }{4}–1,\frac{\alpha }{2}+1,\alpha –1) \because PR\perp (\frac{1}{4},\frac{1}{2},1)$

$\Rightarrow (\frac{\alpha }{4}–1)\frac{1}{4}+\frac{1}{2}(\frac{\alpha }{2}+1)+(\alpha –1)=0$

$\Rightarrow \frac{\alpha }{16}–\frac{1}{4}+\frac{\alpha }{4}+\frac{1}{2}+\alpha –1=0 \Rightarrow \frac{21}{16}\alpha =\frac{3}{4} \Rightarrow \alpha =\frac{4}{7}$

$\therefore R(\frac{1}{7},\frac{2}{7},\frac{4}{7})$

Now, $RP=\sqrt{{(\frac{1}{7}–1)}^2+{(\frac{2}{7}+1)}^2+{(\frac{4}{7}–1)}^2}$

$=\sqrt{\frac{36}{49}+\frac{81}{49}+\frac{9}{49}}=\frac{\sqrt{126}}{7}=\frac{3\sqrt{14}}{7}$.

(2)

Let $\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}=\lambda$

$\Rightarrow x=3\lambda –6, y=2\lambda , z=\lambda –1$

and let $\frac{x–7}{4}=\frac{y–9}{3}=\frac{z–4}{2}=\mu$

$\Rightarrow x=4\mu +7, y=3\mu +9, z=2\mu +4$

Since the given lines intersect so we have

$3\lambda –6=4\mu +7, 2\lambda =3\mu +9, \lambda –1=2\mu +4$

$\Rightarrow 3\lambda –4\mu =13 and \lambda =2\mu +5$

$\Rightarrow 3(2\mu +5)–4\mu =13 \Rightarrow 2\mu =–2$

$\Rightarrow \mu =–1 and \lambda =3$

So, point of intersection is (3, 6, 2)

$\therefore d=\sqrt{{(3–7)}^2+{(6–8)}^2+{(2–9)}^2}$

      $=\sqrt{16+4+49}=\sqrt{69}$

$\therefore d^2+6=69+6=75$.

(2)

We have lines $\frac{2–x}{3}=\frac{3y–2}{4\lambda +1}=4–z$ and $\frac{x+3}{3\mu }=\frac{1–2y}{6}=\frac{5–z}{7}$ are perpendicular.

i.e., $\frac{x–2}{–3}=\frac{y–2/3}{{\displaystyle \frac{4\lambda +1}{3}}}=\frac{z–4}{–1}$ and $\frac{x–(–3)}{3\mu }=\frac{y–1/2}{{\displaystyle \frac{{\displaystyle –6}}{2}}}=\frac{z–5}{–7}$ are perpendicular.

$\Rightarrow –3\left(3\mu \right)+\frac{4\lambda +1}{3}(–3)+(–1)(–7)=0$

$\Rightarrow –9\mu –4\lambda –1+7=0$

$\Rightarrow –9\mu –4\lambda +6=0 \Rightarrow 4\lambda +9\mu =6$.

(2)

Let $\frac{x–1}{2}=\frac{y+1}{3}=\frac{z–2}{5}=\lambda$

$\Rightarrow x=2\lambda +1, y=3\lambda –1 \mathrm{and} z=5\lambda +2$

Clearly, $\overrightarrow{PQ}·(2\hat{i}+3\hat{j}+5\hat{k})=0$

$\Rightarrow (2\lambda –7)·2+(3\lambda –6)·3+(5\lambda –5)·5=0$

$\Rightarrow 38\lambda =57$

$\Rightarrow \lambda =\frac{57}{38}=\frac{3}{2}$

$\therefore$  $(4, \frac{7}{2}, \frac{19}{2})$ are the coordinates of Q.

Now, Q is the midpoint of PP'.

$\therefore \frac{8+\alpha }{2}=4, \frac{5+\beta }{2}=\frac{7}{2}, \frac{7+\gamma }{2}=\frac{19}{2}$

$\Rightarrow (\alpha ,\beta ,\gamma )=(0,2,12)$

$\therefore \alpha +\beta +\gamma =14$.

(3)

$L_1 : \frac{x–3}{2}=\frac{y+15}{–7}=\frac{z–9}{5}$

$L_2 : \frac{x+1}{2}=\frac{y–1}{1}=\frac{z–9}{–3}$

$\overrightarrow{a_1}=3\hat{i}–15\hat{j}+9\hat{k}, \overrightarrow{a_2}=–\hat{i}+\hat{j}+9\hat{k}$

$\overrightarrow{b_1}=2\hat{i}–7\hat{j}+5\hat{k}, \overrightarrow{b_2}=2\hat{i}+\hat{j}–3\hat{k}$

Shortest distance = $=\frac{\left|\right({\overrightarrow{a}}_2-{\overrightarrow{a}}_1)·({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2\left)\right|}{|{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2|}$

$=\frac{\left|\right(–4\hat{i}+16\hat{j})·|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& –7& 5\\ 2& 1& –3\end{array}\left|\right|}{|{\overrightarrow{b}}_1\times {\overrightarrow{b}}_2|}$

$=\frac{\left|\right(–4\hat{i}+16\hat{j})·(16\hat{i}+16\hat{j}+16\hat{k}\left)\right|}{|16\hat{i}+16\hat{j}+16\hat{k}|}=\frac{192}{16\sqrt{3}}=4\sqrt{3}$.

(1)

$L : \frac{x–0}{1}=\frac{y–3}{1}=\frac{z–1}{–1}=\mu$ be the given line.

Any point on L is given by $M(\mu , \mu +3, –\mu +1)$

Direction ratios of $QM=(\mu –3, \mu +6, –\mu )$

Direction ratios of L = (1, 1, –1)

Now, QM $\perp$ L

So, we have, $1·(\mu –3)+1·(\mu +6)–1·(–\mu )=0$

$\Rightarrow 3\mu +3=0$

$\Rightarrow \mu =–1$

So, M(–1, 2, 2) is mid point of PQ.

$\Rightarrow \frac{\alpha +3}{2}=–1, \frac{\beta –3}{2}=2, \frac{\gamma +1}{2}=2$

$\Rightarrow P(\alpha ,\beta ,\gamma )\equiv (–5,7,3)$

Area of $△PQR=\frac{1}{2}|\overrightarrow{PQ}\times \overrightarrow{QR}|$

$\Rightarrow \lambda =\frac{1}{2}$$\left|\right|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 8& –10& –2\\ –1& 8& –2\end{array}\left|\right|$

$\Rightarrow \lambda =\frac{1}{2}$$|36\hat{i}+18\hat{j}+54\hat{k}|$

$\Rightarrow {\lambda }^2=\frac{1}{4}\left(4536\right)$

$\Rightarrow {\lambda }^2=1134=14K \Rightarrow k=\frac{1134}{14}=81$.

(2)

$L_1 : \frac{x–2}{1}=\frac{y–1}{–3}=\frac{z–3}{4}$

$L_2 : \frac{x–2}{2}=\frac{y–3}{3}=\frac{z–5}{1}$

$\therefore a_1=2\hat{i}+\hat{j}+3\hat{k}, a_2=2\hat{i}+3\hat{j}+5\hat{k}$

     $n_1=\hat{i}–3\hat{j}+4\hat{k}, n_2=2\hat{i}+3\hat{j}+\hat{k}$

Shortest distance = $\frac{\left|\right({\overrightarrow{a}}_2–{\overrightarrow{a}}_1)·({\overrightarrow{n}}_1\times {\overrightarrow{n}}_2\left)\right|}{|{\overrightarrow{n}}_1\times {\overrightarrow{n}}_2|}$

$=\frac{\left|\right(2\hat{j}+2\hat{k})·|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& –3& 4\\ 2& 3& 1\end{array}\left|\right|}{|{\overrightarrow{n}}_1\times {\overrightarrow{n}}_2|}$

$=$$\left|\frac{(2\hat{j}+2\hat{k})·(–15\hat{i}+7\hat{j}+9\hat{k})}{\sqrt{{\left(15\right)}^2+7^2+9^2}}\right|$$=\frac{32}{\sqrt{355}}=\frac{m}{\sqrt{n}}$

On comparing, we get m = 32 and n = 355.

So, m + n = 32 + 355 = 387.

(3)

We have $\overrightarrow{a}=\lambda \hat{i}+4\hat{j}+3\hat{k}, \overrightarrow{b}=2\hat{i}+4\hat{j}+7\hat{k}$ and $\overrightarrow{p}=2\hat{i}+3\hat{j}+4\hat{k}$

Shortest distance, $d=$$\left|\frac{(\overrightarrow{b}–\overrightarrow{a})\times \overrightarrow{p}}{\left|\overrightarrow{p}\right|}\right|$

$\overrightarrow{b}–\overrightarrow{a}=(2–\lambda )\hat{i}+4\hat{k}$

$(\overrightarrow{b}–\overrightarrow{a})\times \overrightarrow{p}=$$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2–\lambda & 0& 4\\ 2& 3& 4\end{array}\right|$

     $=\hat{i}(–12)–\hat{j}(8–4\lambda –8)+\hat{k}(6–3\lambda )$

     $=–12\hat{i}+4\lambda \hat{j}+3(2–\lambda )\hat{k}$

$\therefore d=$$\left|\frac{\sqrt{144+16{\lambda }^2+9{(2–\lambda )}^2}}{\sqrt{4+9+16}}\right|$$=\frac{13}{\sqrt{29}}$          (Given)

$\Rightarrow \sqrt{144+16{\lambda }^2+36+9{\lambda }^2–36\lambda }=13$

$\Rightarrow 25{\lambda }^2–36\lambda +180=169 \Rightarrow 25{\lambda }^2–36\lambda +11=0$

$\Rightarrow (25\lambda –11)(\lambda –1)=0 \Rightarrow \lambda =\frac{11}{25} \mathrm{or} 1$.

(3)

Given lines can be written as

$\frac{x–3}{4}=\frac{y–(–7)}{–11}=\frac{z–1}{5}$ and $\frac{x–5}{3}=\frac{y–9}{–6}=\frac{z–(–2)}{1}$

${\overrightarrow{a}}_1=3\hat{i}–7\hat{j}+\hat{k}, {\overrightarrow{a}}_2=5\hat{i}+9\hat{j}–2\hat{k}, {\overrightarrow{b}}_1=4\hat{i}–11\hat{j}+5\hat{k}$

and ${\overrightarrow{b}}_2=3\hat{i}–6\hat{j}+\hat{k}$

$\therefore {\overrightarrow{a}}_2–{\overrightarrow{a}}_1=2\hat{i}+16\hat{j}–3\hat{k}$ and ${\overrightarrow{b}}_1\times {\overrightarrow{b}}_2=$$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 4& –11& 5\\ 3& –6& 1\end{array}\right|$$=19\hat{i}+11\hat{j}+9\hat{k}$

Shortest distance between two lines,

$d=\frac{\left|\right({\overrightarrow{a}}_2–{\overrightarrow{a}}_1)·({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2\left)\right|}{\left|\right({\overrightarrow{b}}_1\times {\overrightarrow{b}}_2\left)\right|} \Rightarrow d=\frac{|38+176–27|}{\sqrt{361+121+81}}=\frac{187}{\sqrt{563}}$.