(2)

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

$L_2 : \frac{x+1}{(1/2)}=\frac{y–1}{(1/2)}=\frac{z+1}{1}=\mu$

Any point on $L_1$ and $L_2$ will be of the form $(\lambda +2,–\lambda ,\lambda +1)$ and $(\frac{\mu }{2}–1,\frac{\mu }{2}+1,\mu –1)$ respectively.

Now direction ratios of line L are given by

$< \lambda –\frac{\mu }{2}+3, –\lambda –\frac{\mu }{2}–1, \lambda –\mu +2>$

Also L is parallel to line $\frac{x–2}{3}=\frac{y–1}{1}=\frac{z–2}{2}$

$\Rightarrow \frac{\lambda –{\displaystyle \frac{\mu }{2}}+3}{3}=\frac{–\lambda –{\displaystyle \frac{\mu }{2}}–1}{1}=\frac{\lambda –{\displaystyle \mu }+2}{2}$

$\begin{array}{c}\Rightarrow \lambda –\frac{\mu }{2}+3=–3\lambda –\frac{3\mu }{2}–3 ... \left(\mathrm{i}\right)\\ 2(\lambda –\frac{\mu }{2}+3)=3(\lambda –\mu +2) ... \left(\mathrm{ii}\right)\end{array}\} \lambda =\frac{–4}{3}, \mu =\frac{–2}{3}$

$\therefore$  Points will be $(\frac{2}{3},\frac{4}{3},\frac{–1}{3})$ and $(\frac{–4}{3},\frac{2}{3},\frac{–5}{3})$

$\therefore$  L is given by $\frac{x–{\displaystyle \frac{2}{3}}}{3}=\frac{y–{\displaystyle \frac{4}{3}}}{1}=\frac{z+{\displaystyle \frac{1}{3}}}{2}$

Point $(\frac{–1}{3},1,–1)$ will satisfy this line L.

(2)

Equation of line L is given by

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

i.e., $\frac{x–1}{1}=\frac{y–2}{1}=\frac{z–3}{2}=\lambda$

$\therefore$  Any point on L is given by $(\lambda +1, \lambda +2, 2\lambda +3)$

Also, any point of $L_1 : \frac{3x–11}{2}=\frac{3y–11}{1}=\frac{3z–19}{2}=\mu$

is given by $(\frac{2\mu +11}{3},\frac{\mu +11}{3},\frac{2\mu +19}{3})$

Now, $\lambda +1=\frac{2\mu +11}{3}, \lambda +2=\frac{\mu +11}{3}, 2\lambda +3=\frac{2\mu +19}{3}$

$\Rightarrow 3\lambda –2\mu =8 \mathrm{and} 3\lambda –\mu =5$

On solving these two equations, we get $\lambda =\frac{2}{3}$ and $\mu =–3$

Point on L is $(\frac{5}{3},\frac{8}{3},\frac{13}{3})$

Required distance = $\sqrt{{(\frac{11}{3}–\frac{5}{3})}^2+{(\frac{11}{3}–\frac{8}{3})}^2+{(\frac{19}{3}–\frac{13}{3})}^2}$

$=\sqrt{4+1+1}=3$ units.

(1)

The shortest distance between the given lines is given by

$d=$$\left|\frac{\left|\begin{array}{ccc}{x}_2–x_1& y_2–y_1& z_2–z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|}{\sqrt{{(a_1{b}_2–a_2{b}_1)}^2+{(b_1{c}_2–b_2{c}_1)}^2+{(c_1{a}_2–c_2{a}_1)}^2}}\right|$

$\Rightarrow 1=$$\left|\frac{\left|\begin{array}{ccc}\sqrt{3}–\lambda & –1& 1\\ –2& 1& 1\\ 1& –2& 1\end{array}\right|}{\sqrt{{\left(3\right)}^2+{\left(3\right)}^2+{\left(3\right)}^2}}\right|$

$\Rightarrow 1=\frac{\left|\right(\sqrt{3}–\lambda \left)\right(3)+1(–3)+1(3\left)\right|}{\sqrt{27}}$

$\Rightarrow 3\sqrt{3}=3\sqrt{3}–3\lambda \mathrm{or} –3\sqrt{3}=3\sqrt{3}–3\lambda$

$\Rightarrow \lambda =0 \mathrm{or} 6\sqrt{3}=3\lambda \Rightarrow \lambda =2\sqrt{3}$

$\therefore$  The sum of possible values of $\lambda =0+2\sqrt{3}=2\sqrt{3}$.

(1)

The given equation of line is,

$\frac{x+3}{8}=\frac{y–4}{2}=\frac{z+1}{2}=\lambda$ (say)

$\therefore$  Any point on line is $(8\lambda –3, 2\lambda +4, 2\lambda –1)$

Now, distance of this point from R(1, 2, 3)

$\Rightarrow {(8\lambda –4)}^2+{(2\lambda +2)}^2+{(2\lambda –4)}^2=36$

$\Rightarrow 72{\lambda }^2–72\lambda =0 \Rightarrow \lambda (\lambda –1)=0$

$\Rightarrow \lambda =0 \mathrm{or} \lambda =1$

So, coordinates of P = (–3, 4, –1) and coordinates of Q = (5, 6, 1).

Now, centroid of $△PQR$ = (1, 4, 1) $\Rightarrow \alpha =1, \beta =4, \gamma =1$

So, ${\alpha }^2+{\beta }^2+{\gamma }^2=18$.

(4)

Given equation of line is

$\frac{x–1}{3}=\frac{y+1}{2}=\frac{z–2}{1}=\lambda$ (say)

$\therefore$  Any point on line, say $m=(3\lambda +1, 2\lambda –1, \lambda +2)$

Now, d.r.'s of  $PM=<3\lambda –2, 2\lambda –5, \lambda –7>$ and d.r.'s of given line = $<3, 2, 1> \because PM\perp \mathrm{line}$

$\therefore 3(3\lambda –2)+2(2\lambda –5)+1(\lambda –7)=0$

$\Rightarrow 14\lambda =23 \Rightarrow \lambda =\frac{23}{14}$

Now, M is the mid point of P(3, 4, 9) and $Q(\alpha ,\beta ,\gamma )$

$\therefore \frac{83}{14}=\frac{3+\alpha }{2} \Rightarrow \alpha =\frac{62}{7} \Rightarrow \frac{16}{7}=\frac{4+\beta }{2} \Rightarrow \beta =\frac{4}{7}$

and  $\frac{51}{14}=\frac{9+\gamma }{2} \Rightarrow \gamma =\frac{–12}{7}$

So, $14(\alpha +\beta +\gamma )=14\times \frac{54}{7}=108$.

(4)

Let A be the point (7, –2, 11).

Equation of line AB is $\frac{x–7}{2}=\frac{y+2}{–3}=\frac{z–11}{6}=\lambda$

$\Rightarrow x=2\lambda +7, y=–3\lambda –2 \mathrm{and} z=6\lambda +11$

Point B is of the form $(2\lambda +7, –3\lambda –2, 6\lambda +11)$.

Now, point B lies on the line $\frac{x–6}{1}=\frac{y–4}{0}=\frac{z–8}{3}$

$\Rightarrow \frac{2\lambda +7–6}{1}=\frac{–3\lambda –2–4}{0}=\frac{6\lambda +11–8}{3}$

$\Rightarrow –3\lambda –6=0 \Rightarrow \lambda =–2$

Thus point B is (3, 4, –1).

$\therefore$  Required distance = $AB=\sqrt{4^2+{(–6)}^2+{12}^2}$

.$=\sqrt{16+36+144}=\sqrt{196}=14 \mathrm{units}$.

(3)

Here $x_1=4, y_1=–1, z_1=0$ and $x_2=\lambda , y_2=–1, z_2=2$

$a_1=1, b_1=2, c_1=–3$ and $a_2=2, b_2=4, c_2=–5$

$\because \frac{6}{\sqrt{5}}=$$\left|\frac{\left|\begin{array}{ccc}\lambda –4& 0& 2\\ 1& 2& –3\\ 2& 4& –5\end{array}\right|}{\sqrt{{(–10+12)}^2+{(–6+5)}^2+{(4–4)}^2}}\right|$

$\Rightarrow \frac{6}{\sqrt{5}}=$$\left|\frac{2\lambda –8}{\sqrt{5}}\right|$$\Rightarrow \lambda =7 \mathrm{or} 1$.

(4)

Let $L_1 : \frac{x}{1}=\frac{y–1}{2}=\frac{z–2}{3}=\lambda$ (say)

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

Let the coordinates of M are $(\lambda ,2\lambda +1,3\lambda +2)$ lie on the given line.

Let given point is P(1, 0, 7).

$\therefore$  Direction ratio of PM are $(\lambda –1,2\lambda +1,3\lambda –5)$ PM is perpendicular the given line $L_1$.

$\therefore 1(\lambda –1)+2(2\lambda +1)+3(3\lambda –5)=0$

$\Rightarrow \lambda –1+4\lambda +2+9\lambda –15=0$

$\Rightarrow 14\lambda –14=0 \Rightarrow \lambda =1$

$\therefore$  The coordinates of M are (1, 3, 5).

$\because$  M is mid point of PQ.

So, $\frac{\alpha +1}{2}=1,\frac{\beta }{2}=3$ and $\frac{\gamma +7}{2}=5$

$\Rightarrow \alpha =1, \beta =6 \mathrm{and} \gamma =3$

$\therefore$  The image of point P(1, 0, 7) is Q(1, 6, 3).

Now the direction cosine of the line are

$m=\frac{\cos 2\pi }{3}=\cos (\pi –\frac{\pi }{3})=–\cos \frac{\pi }{3}=–\frac{1}{2}$

$n=\cos \left(\frac{3\pi }{4}\right)=\cos (\pi –\frac{\pi }{4})=–\frac{1}{\sqrt{2}}$

We know that, $l^2+m^2+n^2=1$

$\Rightarrow l^2+\frac{1}{4}+\frac{1}{2}=1$

$\Rightarrow l^2=1–\frac{1}{4}–\frac{1}{2}=\frac{4–1–2}{4}=\frac{1}{4} \Rightarrow l=\pm \frac{1}{2}$

$\Rightarrow l=\frac{1}{2}$        ($\because$  Line make an acute angle with x-axis)

The equation of line passing through (1, 6, 3) with direction cosines $\frac{1}{2}, –\frac{1}{2}$ and $–\frac{1}{\sqrt{2}}$ is given by

$\overrightarrow{r}=\hat{i}+6\hat{j}+3\hat{k}+\lambda (\frac{1}{2}\hat{i}–\frac{1}{2}\hat{j}–\frac{1}{\sqrt{2}}\hat{k})$

$=(1+\frac{\lambda }{2})\hat{i}+(6–\frac{\lambda }{2})\hat{j}+(3–\frac{\lambda }{\sqrt{2}})\hat{k}$

Only option (4) satisfied the above equation for $\lambda$ = 4.

(1)

Let centroid G divides MR in the ratio 1 : 2.

So, centroid is given by

 $\therefore G(\frac{4–1}{3},\frac{2+4}{3},\frac{4+2}{3})$ i.e., G(1, 2, 2)

Let $l_1 : \frac{x–2}{0}=\frac{y}{2}=\frac{z+3}{–1}=\lambda$ (say)

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

$l_2 : \frac{x–1}{1}=\frac{y+3}{–3}=\frac{z+1}{1}=r$ (say)

$\Rightarrow$ x = r + 1, y = –3r – 3, z = r – 1

Now, $2=r+1 \Rightarrow r=1$ and $2\lambda =–3r–3 \Rightarrow \lambda =–3$.

$\therefore$  Point of intersection of lines $l_1$ and $l_2$ is given by A(2, –6, 0).

$\therefore$  Required distance, $AG=\sqrt{1+64+4}=\sqrt{69}$.

(3)

Direction ratios of line PQ = <4 – 3, 6 – 2, 2 – 3> = <1, 4, –1>

Direction ratios of line PR = <7 – 3, 3 – 2, 2 – 3> = <4, 1, –1>

Let $\theta$ be the required angle.

$\therefore \cos \theta =$$\left|\frac{1\times 4+4\times 1+(–1)\times (–1)}{\sqrt{1+16+1}\sqrt{16+1+1}}\right|$$=$$\left|\frac{9}{18}\right|$$=\frac{1}{2}$

$\Rightarrow \theta ={\cos }^{–1}\left(\frac{1}{2}\right)=\frac{\pi }{3} \therefore \angle QPR=\frac{\pi }{3}$.