(4)

Let the parallel line is $\frac{x–1}{1}=\frac{y–4}{2}=\frac{z–0}{3}=\lambda$

$\Rightarrow x=\lambda +1, y=2\lambda +4, z=3\lambda$

and $\frac{x–2}{2}=\frac{y–6}{3}=\frac{z–3}{4}=t$

$\Rightarrow x=2t+2, y=3t+6, z=4t+3$

So, their point of intersection is

$(\lambda +1,2\lambda +4,3\lambda )=(2t+2,3t+6,4t+3)$

$\Rightarrow \lambda =2t+1$

and $2\lambda +4=3t+6$          [$\because \lambda =2t+1$]

$\Rightarrow t=0, \lambda =1$

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

So, distance $=\sqrt{{(2–1)}^2+{(6–4)}^2+{(3–0)}^2}=\sqrt{14}$.

(1)

Equation of line passing through (–1, 2, 1) and parallel to $\frac{x–1}{2}=\frac{y+1}{3}=\frac{z}{4}$ is $\frac{x+1}{2}=\frac{y–2}{3}=\frac{z–1}{4}$.

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

Coordinates of P are $(2\lambda –1,3\lambda +2,4\lambda +1)$.

As point P also lies on

$\frac{x+2}{3}=\frac{y–3}{2}=\frac{z–4}{1}=\mu \left(\mathrm{say}\right)$

$\therefore$   Coordinates of point P are $(3\mu –2,2\mu +3,\mu +4)$

$\Rightarrow 2\lambda –1=3\mu –2, 3\lambda +2=2\mu +3, 4\lambda +1=\mu +4$

On solving, we get $\lambda =\mu =1$

$\therefore$   The point P is (1, 5, 5)

$\therefore$   Required distance $QP=\sqrt{{(4–1)}^2+{(–5–5)}^2+{(1–5)}^2}$

                                                     $=\sqrt{9+100+16}=5\sqrt{5}$

(4)

Let BM be the height of the triangle ABC.

Direction ratios of AC = 3, 2, –2

Coordinates of $M=(3\lambda +6,2\lambda +7,–2\lambda +7)$

Direction ratios of BM $=(3\lambda +6–1,2\lambda +7–2,–2\lambda +7–3)$

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

$\because BM\perp AC$

$\therefore 3(3\lambda +5)+2(2\lambda +5)–2(–2\lambda +4)=0$

$\Rightarrow 9\lambda +15+4\lambda +10+4\lambda –8=0$

$\Rightarrow 17\lambda +17=0 \Rightarrow \lambda =–1$

$\therefore$   Coordinates of M = (3, 5, 9)

$\therefore BM=\sqrt{{(3–1)}^2+{(5–2)}^2+{(9–3)}^2}=7$

Area of $△ABC==\frac{1}{2}\times 7\times 6=21 \mathrm{sq}. \mathrm{units}$.

(1)

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

$\Rightarrow M\equiv (2\lambda +1,\lambda +2,3\lambda +1)$

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

Since, $\overrightarrow{PM}$ is perpendicular to the given line

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

$\Rightarrow \lambda =1$

$\therefore$   The coordinates of point M is (3, 3, 4).

Let Q be the image of the point P. Then, M be the mid-point of PQ.

$\therefore (\frac{4+\alpha }{2},\frac{4+\beta }{2},\frac{3+\gamma }{2})\equiv (3,3,4)$

$\Rightarrow$ $\mathrm{\alpha }=2,\mathrm{\beta }=2,\mathrm{\gamma }=5$      $\therefore$  $\mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }=9$

(1)

Equation of line passing through the point $(\frac{15}{7},\frac{32}{7},7)$ having direction ratios 1, 4 and 7 is given by

$\frac{x–{\displaystyle \frac{15}{7}}}{1}=\frac{y–{\displaystyle \frac{32}{7}}}{4}=\frac{z–7}{7}=\lambda \left(\mathrm{say}\right)$

$\therefore x=\lambda +\frac{15}{7}, y=4\lambda +\frac{32}{7} \mathrm{and} z=7\lambda +7$

It lies on the given line $\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$

i.e., $\frac{\lambda +{\displaystyle \frac{15}{7}}+1}{3}=\frac{7\lambda +7+5}{7}$

$\Rightarrow 7\lambda +22=21\lambda +36$

$\Rightarrow \lambda =–1$

$\therefore (x,y,z)\equiv (\frac{8}{7},\frac{4}{7},0)$

Square of the distance of the point $(\frac{15}{7},\frac{32}{7},7)$ and $(\frac{8}{7},\frac{4}{7},0)$

$={(\frac{15}{7}–\frac{8}{7})}^2+{(\frac{32}{7}–\frac{4}{7})}^2+{(7–0)}^2$

= 1 + 16 + 49 = 66.

(1)

Equation of line OA is $x–\sqrt{3}y=0$

Equation of line OB is $\sqrt{3}x–y=0$

Equation of angle bisector is $\frac{x–\sqrt{3}y}{\sqrt{1+3}}+\frac{\sqrt{3}x–y}{\sqrt{3+1}}=0$

$\Rightarrow x–y=0$

$\left|\frac{a–(1–a)}{\sqrt{2}}\right|$$=\frac{9}{\sqrt{2}} \Rightarrow a=5 \mathrm{or} –4$

Required sum = 5 + (–4) = 1.

(4)

We have, $L_1:–\hat{i}+2\hat{j}+\hat{k}+\lambda (\hat{i}+2\hat{j}+\hat{k}) \mathrm{and} L_2:\hat{j}+\hat{k}+\mu (2\hat{i}+7\hat{j}+3\hat{k})$

Point of intersection of $L_1$ & $L_2$ is given by

$(–1+\lambda )\hat{i}+(2+2\lambda )\hat{j}+(1+\lambda )\hat{k}=2\mu \hat{i}+(1+7\mu )\hat{j}+(1+3\mu )\hat{k}$

$\Rightarrow –1+\lambda =2\mu , 2+2\lambda =1+7\mu , 1+\lambda =1+3\mu$

$\Rightarrow \mu =1 \mathrm{and} \lambda =3$

$\therefore$   Position vector of point of intersection of $L_1$ and $L_2$ is $2\hat{i}+8\hat{j}+4\hat{k}$.

Hence, $L_3$ is given by

$2\hat{i}+8\hat{j}+4\hat{k}+\gamma (\overrightarrow{a}+\overrightarrow{b})=2\hat{i}+8\hat{j}+4\hat{k}+\gamma (3\hat{i}+9\hat{j}+4\hat{k})$

For $\mathrm{\gamma }=2,$ line ${\mathrm{L}}_3$ passes through point (8, 26, 12).

(2)

Let D.r.s. of $L_3$ be a, b, c

Now, $L_3$ is ${\perp }^r$ to $L_1$ and $L_2$ = a – b + 2c = 0          ... (i)

and –a + 2b + c = 0          ... (ii)

Solving (i) and (ii), we get $\frac{a}{–5}=\frac{b}{–3}=\frac{c}{1}$

Equation of line $L_3$ is $\frac{x–\alpha }{–5}=\frac{y–\beta }{–3}=\frac{z–\gamma }{1}=k$

Any point on $L_3$ is $(–5k+\alpha ,–3k+\beta ,k+\gamma )$

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

Any point on $L_1$ is $(\lambda +1,–\lambda +2,2\lambda +1)$

Now $L_1$ and $L_3$ intersects.

$\therefore –5k+\alpha =\lambda +1, –3k+\beta =–\lambda +2, k+\gamma =2\lambda +1$

$\Rightarrow \alpha =5k+\lambda +1, \beta =3k–\lambda +2, \gamma =–k+2\lambda +1$

$\therefore |5\alpha –11\beta –8\gamma |$

$=|25k+5\lambda +5–33k+11\lambda –22+8k–16\lambda –8|=|–25|=25$.

(4)

Centroid of $△ABC=(\frac{9+3+5}{3},\frac{11+4+13}{3})=(\frac{17}{3},\frac{28}{3})$

Let image of centroid with respect to line mirror is (h, k).

$\frac{h–{\displaystyle \frac{17}{3}}}{3}=\frac{k–{\displaystyle \frac{28}{3}}}{6}=\frac{–2(3\times {\displaystyle \frac{17}{3}}+6\times {\displaystyle \frac{28}{3}}–53)}{3^2+6^2}$

Solving, we get h = 3 and k = 4

$\therefore h^2+k^2+hk=37$.

(3)

We have, $L:\frac{x–1}{1}=\frac{y+1}{–1}=\frac{z–2}{2}=\mu$

$\Rightarrow \mathrm{Point} \mathrm{is} P(\mu +1,–\mu –1,2\mu +2)$

$\therefore \mathrm{D}.\mathrm{r}\text{'}\mathrm{s} \mathrm{of} AP=(\mu ,–\mu –3,2\mu )$

Now, $\mu \left(1\right)+(–\mu –3)(–1)+2\left(2\mu \right)=0$

$\Rightarrow \mu +\mu +3+4\mu =0$

$\Rightarrow 6\mu +3=0 \Rightarrow \mu =\frac{–1}{2}$

$\therefore P\equiv (\frac{1}{2},\frac{–1}{2},1)$

Now, line L intersect the other line at Q, i.e.,

$(–1+\lambda ,1–\lambda ,–2+\lambda )$

$\therefore \mu +1=–1+\lambda , –\mu –1=1–\lambda \mathrm{and} 2\mu +2=–2+\lambda$

$\Rightarrow \mu =\lambda –2, \mu =\lambda –2 \mathrm{and} 2\mu +2=–2+\lambda$

$\Rightarrow 2(\lambda –2)+2=–2+\lambda$

$\Rightarrow 2\lambda –4+2=–2+\lambda \Rightarrow \lambda =0 \mathrm{and} \mu =–2$

$\therefore Q\equiv (–1,1,–2)$

So, ${\left(PQ\right)}^2={\left(\frac{–3}{2}\right)}^2+{\left(\frac{3}{2}\right)}^2+{(–3)}^2=\frac{9}{4}+\frac{9}{4}+9=9+\frac{9}{2}=\frac{27}{2}$

Hence, $2P{Q}^2=27$.