Three Dimensional Geometry

$ 0=\frac{6+3 \alpha}{2+3} \Rightarrow \alpha=-2 $

$ \begin{aligned} &0=\frac{4 \beta+20}{9} \Rightarrow \beta=-5\\ &\text { }\\ &\begin{aligned} So,\,\,& A \equiv(-2,4,7) \\ & \beta \equiv(3,-5,8) \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \frac{A C}{B C}=\frac{2}{5} \\ & C_x=\frac{2 \times 3-5 \times(-2)}{2-5}=\frac{-16}{3} \\ & C_y=\frac{-10-20}{2-5}=10 \\ & C_z=\frac{2 \times 8-5 \times 7}{-3}=\frac{19}{3} \end{aligned}\\ &\text { So, } \quad C \equiv\left(-\frac{16}{3}, 10, \frac{19}{3}\right) \end{aligned} $

Let direction cosine of $X$-axis be ( $1,0,0$ )

direction cosine of line having direction ratio as $3,-1,5$

$ 1= \pm \frac{3}{\sqrt{35}}, m=\mp-\frac{1}{\sqrt{35}}, n= \pm \frac{5}{\sqrt{35}} $

Direction cosine of line bisecting the angle below two given lines is

$ \begin{aligned} \left(l_1 \pm l_2, m_1 \pm m_2, n_1 \pm n_2\right) & \\ \left(\frac{\sqrt{35}-3}{\sqrt{35}}, \frac{1}{\sqrt{35}}, \frac{-5}{\sqrt{35}}\right) \\ \left(\frac{\sqrt{35}-3}{\sqrt{5}}, \frac{1}{\sqrt{5}},-\sqrt{5}\right) \end{aligned} $

Equation of plane $\pi_1$ :

$ 2 x+y-z-\frac{\alpha}{2}=0 $

Equation of plane $\pi_2: 2 x+y-z+1=0$

Distance below plane $\pi_1$ and plane $\pi_2$

$ \begin{aligned} & =\frac{\left|1+\frac{\alpha}{2}\right|}{\sqrt{2^2+1^2+1^2}}=2=\left|1+\frac{\alpha}{2}\right|=2 \sqrt{6} \\ & \Rightarrow \frac{\alpha}{2}+1= \pm 2 \sqrt{6} \\ & \Rightarrow \alpha+2= \pm 4 \sqrt{6} \\ & \Rightarrow \alpha=-2+4 \sqrt{6} \text { and } \alpha=-2-4 \sqrt{6} \\ & \text { Product }=(-2+4 \sqrt{6})(-2-4 \sqrt{6}) \\ & =4-96=-92 \end{aligned} $

Let coordinates of point be $(x, y, z)$.

Distance from $X Y$ plane $=2 \times$ Distance from $Z$ axis.

$ \begin{aligned} & |z|=2 \times \sqrt{x^2+y^2} \\ \Rightarrow & z^2=4 x^2+4 y^2 \Rightarrow 4 x^2+4 y^2-z^2=0 \end{aligned} $

Let side of cube be $a$.

Let coordinates of vertices of cube are shown as in figure.

Ratio direction ratio of line $A Q$ is $a: a:-a$

Ratio of direction ratio of line $B R$ is $a:-a: a$.

$ \cos \alpha=\frac{\left|a^2-a^2-a^2\right|}{\sqrt{3} a \times \sqrt{3} a}=\frac{1}{3} $

Let ratio of direction ratio of line $B D$ be $a:-a: 0$

$ \begin{array}{ll} & \cos \beta=\frac{a^2+a^2}{\sqrt{3} a \times \sqrt{2} a}=\sqrt{\frac{2}{3}} \\ \Rightarrow & \cos \alpha+\cos ^2 \beta=\frac{1}{3}+\frac{2}{3}=1 \end{array} $

Let direction ratio of normal line of plane $\pi_1$ be $(a,-1,3)$.

Let direction ratio of normal line of plane $\pi_2$ be $(3, a, 1)$

$ \begin{aligned} & \cos 60^{\circ}=\frac{3 a-a+3}{\sqrt{a^2+10} \times \sqrt{a^2+10}}=\frac{2 a+3}{a^2+10} \\ & \Rightarrow \quad \frac{1}{2}=\frac{2 a+3}{a^2+10} \Rightarrow a^2+10=4 a+6 \\ & \Rightarrow \quad a^2-4 a+4=0 \Rightarrow a=2 \end{aligned} $

⇒ Direction ratio of line perpendicular to the plane $(a+2) x+(a-4) y+2 a z=a$

$ 4:-2: 4 $

Direction ratio 2: $-1: 2$

We have,

$ \begin{gathered} A(-4,9, k), B(-1,6, k) \text { and } C(0,7,10) \\ \mathbf{A B}=3 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+0 \hat{\mathbf{k}},|\mathbf{A B}|=3 \sqrt{2} \\ \mathbf{B C}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+(10-k) \hat{\mathbf{k}} \\ |\mathbf{B C}|=\sqrt{1+1+(10-k)^2} \end{gathered} $

And $\mathbf{A C}=4 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+(10-k) \hat{\mathbf{k}}$,

$ \begin{aligned} & |\mathbf{A C}|=\sqrt{16+4+(10-k)^2} \\ \therefore & |\mathbf{A C}|>|\mathbf{B C}| \\ \therefore & A C \rightarrow \text { Hypotenuse } \\ \therefore & \mathbf{A B} \cdot \mathbf{B C}=0 \\ \therefore & |\mathbf{A B}|=|\mathbf{B C}| \\ \Rightarrow & 18=2+(10-k)^2 \\ & 10-k= \pm 4 \Rightarrow k=10 \pm 4=14 \text { or } 6 \end{aligned} $

∴ No. of values of $k=2$

$ \begin{aligned} &\text { We have, }\\ &\begin{array}{ll} & \alpha=60^{\circ}, \beta=45^{\circ}, \gamma=\theta \\ \because & \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1 \\ \Rightarrow & \cos ^2 60+\cos ^2 45^{\circ}+\cos ^2 \theta=1 \\ \Rightarrow & \cos ^2 \theta=1-\frac{1}{4}-\frac{1}{2}=\frac{1}{4} \\ \Rightarrow & \cos \theta=\frac{1}{2} \Rightarrow \tan \theta=\sqrt{3} \end{array} \end{aligned} $

$ \begin{aligned} \mathbf{P Q}= & -\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\ & \pi: a x+b y-3 z+d=0 \end{aligned} $

$ \begin{aligned} &\begin{array}{ll} \therefore & \mathbf{n}=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}-3 \hat{\mathbf{k}} \\ \therefore & \mathbf{P Q} \| \mathbf{n} \\ \therefore & \frac{-1}{a}=\frac{-2}{b}=\frac{3}{-3} \Rightarrow a=1, b=2 \end{array}\\ &\therefore Q(1,-2,0) \text { lies on plane }\\ &\begin{gathered} x+2 y-3 z+d, 1-4+d=0 \\ d=3 \\ a+b+d=1+2+3=6 \end{gathered} \end{aligned} $

For $\pi_1, n_1=a \times b$

where, $\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}$

$ \begin{aligned} \mathbf{b} & =\hat{\mathbf{i}}+\hat{\mathbf{k}} \\ \therefore \mathbf{n}_1 & =\left|\begin{array}{lll} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & 0 \\ 1 & 0 & 1 \end{array}\right|=\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}} \end{aligned} $

For $\pi_2, \mathbf{n}_2=\mathbf{c} \times \mathbf{d}$ where $\mathbf{c}=\hat{\mathbf{j}}-\hat{\mathbf{k}}$

$ \begin{aligned} \mathbf{d} & =\hat{\mathbf{k}}-\hat{\mathbf{i}} \\ \mathbf{n}_2 & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 0 & 1 & -1 \\ -1 & 0 & 1 \end{array}\right|=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \end{aligned} $

Parallel vector line of intersection of $\pi_1$ and $\pi_2$

$ \begin{aligned} & \mathbf{a}=\mathbf{n}_1 \times \mathbf{n}_2 \\ & \mathbf{n}_2=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -1 & -1 \\ 1 & 1 & 1 \end{array}\right| \\ & \quad=\hat{\mathbf{i}}(-1+1)-\hat{\mathbf{j}}(1+1)+\hat{\mathbf{k}}(1+1) \\ & \quad=-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\ & \begin{aligned} \text { And } \mathbf{b} & =\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}} \\ \therefore \cos \theta & =\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|} \\ \quad & =\left|\frac{-2-2}{\sqrt{4+4} \sqrt{3}}\right|=\left|\frac{4}{2 \sqrt{2} \sqrt{3}}\right| \\ \theta & =\cos ^{-1} \sqrt{\frac{2}{3}} . \end{aligned} \end{aligned} $

$ \text { Let } \frac{m}{n}=\frac{\lambda}{1} $

$ \begin{aligned} & \frac{p \lambda+2}{\lambda+1}=\frac{8}{5} \\ & 5 p \lambda+10=8 \lambda+8 \end{aligned} $

$ \begin{aligned} &\begin{gathered} 5 p \lambda-8 \lambda=-2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\ \lambda(8-5 p)=2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \frac{-2 \lambda+p}{\lambda+1}=\frac{-1}{5} \\ -10 \lambda+5 p=-\lambda-1 \Rightarrow 5 p+1=9 \lambda \\ \lambda=\frac{5 p+1}{9}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \\ \frac{\lambda p+2}{\lambda+1}=\frac{8}{5} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{gathered}\\ &\text { Solving Eqs. (i) and (ii), we get }\\ &\begin{array}{ll} & \left(\frac{5 p+1}{9}\right)(8-5 p)=2 \\ \Rightarrow & (5 p+1)(8-5 p)=18 \\ \Rightarrow & 40 p-25 p^2+8-5 p-18=0 \\ \Rightarrow & -25 p^2+35 p-10=0 \\ \Rightarrow & 5 p^2-7 p+2=0 \\ \therefore & p=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\because p \in Z)\\ & \lambda=\frac{5+1}{9}=\frac{6}{9}=\frac{2}{3} . \\ \because & \frac{m}{n}=\lambda \therefore \frac{3 m+n}{3 n}=\frac{m}{n}+\frac{1}{3} \\ & =\frac{2}{3}+\frac{1}{3}=1=p \end{array} \end{aligned} $

Let $M$ be the foot of perpendicular

$ M=\left(\frac{\lambda+2}{\lambda+1}, \frac{\lambda-1}{\lambda+1}, \frac{-2 \lambda+1}{\lambda+1}\right) $

$\because \mathbf{P M} \cdot \mathbf{A B}=0$

$ \begin{aligned} & \left(\frac{\lambda+2}{\lambda+1}+1\right)(1)+\left(\frac{\lambda-1}{\lambda+1}-2\right)(-2) +\left(\frac{-2 \lambda+1}{\lambda+1}+1\right) 3=0 \\ & \begin{aligned} & \therefore 2 \lambda+3-2(-\lambda-3)+3(-\lambda+2)=0 \\ & \Rightarrow 2 \lambda+3+2 \lambda+6-3 \lambda+6=0 \Rightarrow \lambda=-15 \\ & \therefore \alpha+\beta+\gamma=\frac{\lambda+2+\lambda-1-2 \lambda+1}{\lambda+1} \\ &= \frac{2}{\lambda+1}=\frac{2}{-15+1}=\frac{-1}{7} \end{aligned} \end{aligned} $

Equation of plane containing $A(2,1,-1), B(6,-3,2) C(-3,12,4)$ is

$ \begin{aligned} & \left|\begin{array}{ccc} x-2 & y-1 & z+1 \\ 4 & -4 & 3 \\ 9 & -15 & -2 \end{array}\right|=0 \\ & \begin{array}{r} \Rightarrow(x-2)(8+45)-(y-1)(-8-27) +(z+1)(-60+36)=0 \end{array} \\ & \Rightarrow 53 x-106+35 y-35-24 z-24=0 \\ & \Rightarrow 53 x+35 y-24 z-165=0 \\ & \begin{aligned} \therefore & \frac{d}{b+c}=\frac{-165}{35-24}=\frac{-165}{11}=-15 \end{aligned} \end{aligned} $

Given, normal vector to the plane is $\mathbf{n}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$

And direction vector to the line is $\mathbf{b}=\mathbf{j}-2 \hat{\mathbf{k}}$

Since, $\mathbf{b} \cdot \mathbf{n}=(\hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \cdot(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+5 \hat{\mathbf{k}})$

$ =3-10=-7 \neq 0 $

$\Rightarrow$ line is not perpendicular to the plane Now, $\mathbf{P}(t)=\hat{\mathbf{i}}+(-3+t) \hat{\mathbf{j}}-2 t \hat{\mathbf{k}}$

for minimum distance, $\mathbf{p}(t) \cdot \mathbf{n}=0$

$ \begin{array}{ll} \Rightarrow & 2+(-9+3 t)-10 t=0 \\ \Rightarrow & -7 t-7=0 \Rightarrow t=-1 \\ & \therefore \quad \mathbf{p}=\hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\ \therefore & \quad A P=(\hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})-(\hat{\mathbf{i}}-3 \hat{\mathbf{j}}) \\ & \quad=-\hat{\mathbf{j}}+2 \hat{\mathbf{k}} \end{array} $

$\therefore$ Equation of plane

$ \begin{array}{rr} & (\mathbf{r}-(\hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})) \cdot(-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})=0 \\ \Rightarrow \quad & ((x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}})-(\hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})) \\ & ((-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})=0 \\ \Rightarrow \quad & ((x-1) \hat{\mathbf{i}}+(y+4) \hat{\mathbf{j}}+(z-2 \hat{\mathbf{k}}) \\ & (-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})=0 \end{array} $

$ \begin{array}{ll} \Rightarrow & -(y+4)+2(z-2)=0 \\ \Rightarrow & -y-4+2 z-4=0 \\ \Rightarrow & -y+2 z-8=0 \\ \Rightarrow & -y+2 z=8 \\ \Rightarrow & \mathbf{r} \cdot(-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})=8 \end{array} $

Given, equation of planes

$ 3 x+4 y+7 z=1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

And $\quad x-y+z=5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

$\therefore$ Direction ratio of normal to the plane (i) is $3,4,7$.

and direction ratio of normal to the plane (ii) is $1,-1,1$

$\therefore$ Direction ratio of $L$,

$ \begin{aligned} \mathbf{n} & =\mathbf{n}_1 \times \mathbf{n}_2 \\ & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 3 & 4 & 7 \\ 1 & -1 & 1 \end{array}\right| \\ & =\hat{\mathbf{i}}(4+7)-\hat{\mathbf{j}}(3-7)+\hat{\mathbf{k}}(-3-4) \\ & =11 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-7 \hat{\mathbf{k}} \end{aligned} $

$\therefore$ Direction ratio of $L$ are $11,4,-7$

Given, points $(1,1, \lambda)$ and $(-3,0,1)$ are equidistant from the plane

$ \begin{aligned} & 3 x+4 y-12 z+13=0 \\ & \therefore \quad\left|\frac{3(1)+4(1)-12(\lambda)+13}{\sqrt{9+16+144}}\right| \\ & \quad=\left|\frac{3(-3)+4(0)-12(1)+13}{\sqrt{9+16+144}}\right| \\ & \Rightarrow \quad \frac{|20-12 \lambda|}{\sqrt{169}}=\frac{|-8|}{\sqrt{169}} \\ & \Rightarrow \quad|20-12 \lambda|=8 \end{aligned} $

Taking positive sign, we get

$ \begin{aligned} & 20-12 \lambda=8 \\ \Rightarrow \quad & -12 \lambda=-12 \Rightarrow \lambda=1 \end{aligned} $

And taking negative sign, we get

$ \begin{array}{rlrl} \Rightarrow & & 20-12 \lambda & =-8 \Rightarrow-12 \lambda=-28 \\ \Rightarrow & & 12 \lambda & =28 \Rightarrow \lambda=\frac{7}{3} \\ & \therefore & \lambda =1, \frac{7}{3} \end{array} $

Given,

$ \mathbf{r}_1=(3 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})+t(4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}) $

So, $\mathbf{a}_1=(3 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})$,

$ \mathbf{b}_1=(4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}) $

And $\mathbf{r}_2=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}})+s(6 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})$

So, $\mathbf{a}_2=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}), \mathbf{b}_2=(6 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})$

Since, shortest distance,

$ d=\frac{\left|\left(\mathbf{a}_2-\mathbf{a}_1\right) \cdot\left(\mathbf{b}_1 \times \mathbf{b}_2\right)\right|}{\left|\mathbf{b}_1 \times \mathbf{b}_2\right|} $

Now, $\mathbf{a}_2-\mathbf{a}_1=\langle 1,2,-4\rangle-\langle 3,-5,2\rangle$

$ =\langle 1-3,2-(-5),-4-2\rangle $

$ \begin{aligned} &\begin{gathered} =<-2,7,-6> \\ \mathbf{b}_1 \times \mathbf{b}_2=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 4 & 3 & -1 \\ 6 & 3 & -2 \end{array}\right| \\ =(-6-(-3)) \hat{\mathbf{i}}-(-8-(-6)) \hat{\mathbf{j}}+(12-18) \hat{\mathbf{k}} \\ =-3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-6 \hat{\mathbf{k}} \\ \left|\mathbf{b}_1 \times \mathbf{b}_2\right|=\sqrt{(-3)^2}+2^2+(-6)^2 \\ =\sqrt{9+4+36}=\sqrt{49}=7 \end{gathered}\\ &\text { So, } \quad d=\frac{|\langle-2,7,-6\rangle \cdot\langle-3,2,-6\rangle|}{7}\\ &=\frac{|6+14+36|}{7}=\frac{56}{7}=8 \end{aligned} $

$ \text { } \begin{aligned} A B & =\sqrt{(1-0)^2+(5-3)^2+(6-4)^2} \\ & =\sqrt{1+4+4}=\sqrt{9}=3 \\ A C= & \sqrt{(-2-0)^2+(0-3)^2+(-2-4)^2} \\ = & \sqrt{4+9+36}=\sqrt{49}=7 \end{aligned} $

Using the angle bisector theorem,

$ \frac{B D}{D C}=\frac{A B}{A C}=\frac{3}{7} $

Since, $D$ divides $B C$ in the ratio $3: 7$, using the section formula

$ \begin{aligned} & D=\left(\frac{7 \cdot 1+3 \cdot(-2)}{7+3}, \frac{7 \cdot 5+3 \cdot 0}{7+3}, \frac{7 \cdot 6+3 \cdot(-4)}{7+3}\right) \\ & \Rightarrow\left(\frac{7-6}{10}, \frac{35}{10}, \frac{42-6}{10}\right) \\ & \Rightarrow\left(\frac{1}{10}, \frac{35}{10}, \frac{36}{10}\right)=(0.1,35,3.6) \end{aligned} $

Now,

$ \begin{aligned} & A D=\sqrt{(0.1-0)^2+(3.5-3)^2+(3.6-4)^2} \\ & \Rightarrow \sqrt{(0.1)^2+(0.5)^2+(-0.4)^2} \\ & \Rightarrow \sqrt{0.01+0.25+0.16}=\sqrt{0.42} \\ & \Rightarrow \sqrt{\frac{42}{100}}=\frac{\sqrt{42}}{10} \end{aligned} $

$ \begin{aligned} &\text { Given two lines } 2 l+m-n=0\\ &\begin{aligned} & \Rightarrow \quad n=2 l+m \text { and } l^2-2 m^2+n^2=0 \\ & \Rightarrow \quad l^2-2 m^2+(2 l+m)^2=0 \\ & \Rightarrow \quad l^2-2 m^2+4 l^2+4 l m+m^2=0 \\ & \Rightarrow \quad 5 l^2+4 l m-m^2=0 \\ & \Rightarrow \quad 5\left(\frac{l}{m}\right)^2+4\left(\frac{l}{m}\right)-1= \\ & \text { Let } x=\frac{l}{m}, 5 x^2+4 x-1=0 \\ & \Rightarrow \quad x=\frac{-4 \pm \sqrt{4^2-4(9)(-1)}}{2(9)} \\ & \Rightarrow \quad \frac{-4 \pm \sqrt{36}}{10}=\frac{-4 \pm 6}{10} \\ & \therefore \quad x_1=\frac{2}{10} \text { and } x_2=\frac{-10}{10}=-1 \end{aligned} \end{aligned} $

Case $\mathrm{I} \frac{l}{m}=\frac{1}{5}$

$ \Rightarrow \quad m=5 l $

So, $n=2 l+m=2 l+5 l=7 l$

Direction cosine ( $l_1, m_1, n_1$ ) are proportional to $(l, 5 l, 7 l)$, i.e., $(1,5,7)$

Case II $\frac{l}{m}=-1$

$ \Rightarrow \quad m=-l $

So, $n=2 l+m=2 l-l=l$

Direction cosine ( $l_2, m_2, n_2$ ) are proportional to $(l,-l, l)$ i.e., $(1,-1,1)$

Now,

$ \begin{aligned} & \cos \theta=\frac{l_1 l_2+m_1 m_2+n_1 n_2}{\sqrt{l_1^2+m_1^2+n_1^2} \cdot \sqrt{l_2^2+m_2^2+n_2^2}} \\ & \Rightarrow \frac{1 \times 1+5 \times(-1)+7 \times 1}{\sqrt{1^2+5^2+7^2} \cdot \sqrt{1^2+(-1)^2+1^2}} \\ & \Rightarrow \frac{1-5+7}{\sqrt{1+25+49} \cdot \sqrt{1+1+!}} \\ & \Rightarrow \frac{3}{\sqrt{75} \cdot \sqrt{3}}=\frac{3}{5 \cdot 3} \Rightarrow \frac{3}{15}=\frac{1}{5} \\ & \therefore \cos \theta=\frac{1}{5} \end{aligned} $

Let $P_1=(2,1,2)$ and $P_2=(1,2,1)$

So, $\mathbf{P}_1 \mathbf{P}_2=P_2-P_1$

$ \begin{array}{ll} \Rightarrow & (1,2,1)-(2,1,2) \\ \Rightarrow & (1-2,2-1,1-2 \\ \Rightarrow & (-1,1,-1) \end{array} $

The plane $2 x-y+2 z=1$ has a normal vector $\mathbf{n}_1=(2,-1,2)$.

Since, the required plane is perpendicular to $2 x-y+2 z=1$, its normal vector $\mathbf{n}=(a, b, c)$ must be perpendicular to $\mathbf{n}_1$.

So, $\mathbf{n}$ must be perpendicular to $\mathbf{P}_1 \mathbf{P}_2$.

$\therefore \mathbf{n}$ is parallel to the cross product of $\mathbf{P}_1 \mathbf{P}_2$ and $\mathbf{n}_1$.

$ \begin{aligned} \mathbf{n} & =\mathbf{P}_1 \mathbf{P}_2 \times \mathbf{n}_1=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ -1 & 1 & -1 \\ 2 & -1 & 2 \end{array}\right| \\ & =(2-1) \hat{\mathbf{i}}-(-2+2 \hat{\mathbf{j}}+(1-2 \hat{\mathbf{k}} \\ & =\hat{\mathbf{i}}-\hat{\mathbf{k}}=\langle 1,0,-1\rangle \end{aligned} $

Now, equation of the point $(2,1,2)$ and the normal vector $(1,0,-1)$ is

$ \begin{aligned} & 1(x-2+0(y-1)-1(z-2)=0 \\ & \Rightarrow \quad x-2-z+2=0 \\ & \Rightarrow \quad x-z=0 \end{aligned} $

Comparing this with $a x+b y+c z+d=0$, we have

$ \begin{aligned} & a=1, b=0, c=-1, d=0 \\ \therefore & \frac{a+b}{c+d}=\frac{1+0}{-1+0}=-1 \end{aligned} $

To find the equation of the plane $ \pi $ that passes through the points $ (2, -3, 0) $ and $ (3, 0, 4) $, and is parallel to the vector $ \langle 2, 3, -4 \rangle $, we start with the general equation of a plane:

$ a(x - 2) + b(y + 3) + c(z - 0) = 0. $

Substituting the point $ (3, 0, 4) $ into the plane equation gives:

$ a(1) + b(3) + c(4) = 0 \quad \text{(i)}.$

Since the plane is parallel to the vector $ \langle 2, 3, -4 \rangle $, the plane's normal vector must be orthogonal to $ \langle 2, 3, -4 \rangle $. Thus,

$ 2a + 3b - 4c = 0 \quad \text{(ii)}.$

To find $ a, b, c $ such that both conditions are satisfied, compare ratios:

$ \frac{a}{-12} = \frac{-b}{-4} = \frac{c}{3}. $

Simplifying gives:

$ \frac{a}{-24} = \frac{b}{12} = \frac{c}{-3}. $

From this, solve for proportional constants:

$ \frac{a}{8} = \frac{b}{-4} = \frac{c}{1}. $

Choosing $ c = 1 $, we find $ a = 8 $ and $ b = -4 $.

The equation of the plane becomes:

$ 8(x - 2) - 4(y + 3) + z = 0, $

or simplified:

$ 8x - 4y + z - 28 = 0. $

Next, find the line equation that passes through points $ (1, 2, 0) $ and $ (0, 1, -2) $. The direction vector of the line is $ \langle -1, -1, -2 \rangle $.

The parametric equation of the line is:

$ \frac{x-1}{-1} = \frac{y-2}{-1} = \frac{z-0}{-2} = \lambda. $

Therefore, the line is:

$ P(x, y, z) = (-\lambda + 1, -\lambda + 2, -2\lambda). $

Substitute these into the plane equation:

$ 8(-\lambda + 1) - 4(-\lambda + 2) + (-2\lambda) - 28 = 0. $

Simplifying:

$ -8\lambda + 8 + 4\lambda - 8 - 2\lambda - 28 = 0, $

$ -6\lambda = 28 \quad \Rightarrow \quad \lambda = -\frac{14}{3}. $

Substituting $ \lambda = -\frac{14}{3} $ into the line equations gives the intersection point $ P(a, b, c) = \left(\frac{17}{3}, \frac{20}{3}, \frac{28}{3}\right)$.

Finally, $ a + b + 2c $ is:

$ a + b + 2c = \frac{17}{3} + \frac{20}{3} + \frac{56}{3} = \frac{93}{3} = 31. $

Equation of plane passing through the point of intersection of $ x-y+z-5=0 $ and $ 2 x+y-z-3=0 $ is $ x-y+z-5+\lambda(2 x+y-z-3)=0 $ $ \pi $ is passing through $ (0,1,2) $.

$ \begin{array}{l} 0-1+2-5+\lambda(0+1-2-3)=0 \\\\ \Rightarrow \quad-4+\lambda(-4)=0 \\\\ \Rightarrow \quad-4-4 \lambda=0 \\\\ \Rightarrow \quad 1+\lambda=0 \\\\ \Rightarrow \quad \lambda=-1 \\\\ \Rightarrow x-y+z-5-1(2 x+y-z-3)=0 \\\\ -x-2 y+2 z-2=0 \\\\ x+2 y-2 z+2=0 \\\\ r \cdot(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})=m \\\\ a=1, b=2, c=-2 \\\\ \because \quad \frac{b c}{a^{2}}=-\frac{4}{1} \\\\ \therefore \quad \frac{b c}{a^{2}}=-4 \end{array} $

$ \mathrm{AB}=\mathrm{OB}-\mathrm{OA} =3 \hat{\mathbf{i}}+(b-4) \hat{\mathbf{j}}+(3-a) \hat{\mathbf{k}} $

$ \mathbf{C D}=\mathbf{O D}-\mathbf{O C}=(-5-C) \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} $

$ \mathbf{A B} $ and $ \mathbf{C D} $ are collinear

$ \frac{3}{-5-C}=\frac{b-4}{6}=\frac{3-a}{-3}=\lambda $

$ -5-C=\frac{3}{\lambda}, b-4=6 \lambda $ and $ 3-a=-3 \lambda $

$ c=-5-\frac{3}{\lambda}, b=6 \lambda+4 $ and $ a=3+3 \lambda $

$ a+b+c=-5-\frac{3}{\lambda}+6 \lambda+4+3+3 \lambda $

$ a+b+c=2+9 \lambda-\frac{3}{\lambda} $

By intersection if we put $ \lambda=1 $ then, $ a+b+c=2+9-3=8 $

Direction ratio of

$ \begin{aligned} \mathrm{AB} \equiv(2-1,-1+2,2-1) =(1,1,1) \end{aligned} $

Equation of $ A B $

$ \begin{array}{l} \frac{x-1}{1}=\frac{y+2}{1}=\frac{z-1}{1}=D(\text { say }) \\\\ D \equiv(\lambda+1, \lambda-2, \lambda+1)=D(\alpha, \beta, \gamma) \\\\ \alpha=\lambda+1, \beta=\lambda-2, \gamma=\lambda+1 \end{array} $

Direction ratio of $ C D=\alpha-1, \beta-2, \gamma-3 $ $ A B \perp C D $

$ \alpha-1+\beta-2+\gamma-3=0 $

$ \alpha+\beta+\gamma=6 $

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

$ \alpha=3, \beta=0, \gamma=3 $

$ \alpha^{2}+\beta^{2}+\gamma^{2}=9+0+9=18 $

Direction ratio of $ P M=3,1,-5 $

Let equation of plane passing through

$ \begin{array}{l} \left(x_{1}, y_{1}, z_{1}\right) \\\\ 3\left(x-x_{1}\right)+1\left(y-y_{1}\right)-5\left(z-z_{1}\right)=0 \\\\ 3 x+y-5 z=3 x_{1}+y_{1}-5 z_{1} \end{array} $

$ M(1,0,-2) $ lies on the plane

$ \begin{aligned} & 13=3 x_{1}+y_{1}-5 z_{1} \\\\ \Rightarrow \quad & 3 x+y-5 z=13 \end{aligned} $

intercept on axis is

$ \begin{gathered} \frac{13}{3}, 13, \frac{-13}{5} \\\\ 3 a+b+5 c=13+13-13=13 \end{gathered} $

To find the unit vector normal to the plane π, we first calculate the cross product of the vectors $\hat{\mathbf{i}}+3 \hat{\mathbf{k}}$ and $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$. The cross product is calculated as follows:

$ \mathbf{n} = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 0 & 3 \\ 2 & 1 & -1 \end{vmatrix} = (-3)\hat{\mathbf{i}} + 7\hat{\mathbf{j}} + \hat{\mathbf{k}} $

Now, we find the unit vector $\mathbf{n}_{\text{unit}}$ by dividing this normal vector by its magnitude. The magnitude of $-3\hat{\mathbf{i}} + 7\hat{\mathbf{j}} + \hat{\mathbf{k}}$ is:

$ \sqrt{(-3)^2 + 7^2 + 1^2} = \sqrt{9 + 49 + 1} = \sqrt{59} $

Thus, the unit vector is:

$ \mathbf{n}_{\text{unit}} = \frac{1}{\sqrt{59}}(-3 \hat{\mathbf{i}} + 7 \hat{\mathbf{j}} + \hat{\mathbf{k}}) $

The equation of the plane is given by:

$ -3x + 7y + z + d = 0 $

Since the plane passes through the point $(-3, 7, 1)$, we substitute these coordinates into the plane’s equation:

$ -3(-3) + 7(7) + 1 + d = 0 \\ 9 + 49 + 1 + d = 0 \\ d = -59 $

Therefore, the equation of the plane is:

$ -3x + 7y + z - 59 = 0 $

The perpendicular distance $p$ from the origin to the plane can be calculated using the formula for the distance from a point to a plane:

$ p = \frac{|0(-3) + 0(7) + 0(1) - 59|}{\sqrt{(-3)^2 + 7^2 + 1^2}} = \frac{59}{\sqrt{59}} = \sqrt{59} $

Given $p$, we calculate:

$ \sqrt{p^2 + 5} = \sqrt{(\sqrt{59})^2 + 5} = \sqrt{59 + 5} = \sqrt{64} = 8 $

To find the harmonic conjugate $ Q(\alpha, \beta, \gamma) $ of the point $ P(2,3,4) $ with respect to the line segment joining points $ A(3,-2,2) $ and $ B(6,-17,-4) $, we start by noting that $ P $ divides $ AB $ in some ratio $ \lambda : 1 $.

First, let's determine this ratio using the x-coordinate:

$ 2 = \frac{6\lambda + 3}{\lambda + 1} $

Solving for $ \lambda $:

$ 2(\lambda + 1) = 6\lambda + 3 \\ 2\lambda + 2 = 6\lambda + 3 \\ 2\lambda - 6\lambda = 3 - 2 \\ -4\lambda = 1 \\ \lambda = -\frac{1}{4} $

This indicates that $ P $ divides $ AB $ in the ratio $-\frac{1}{4} : 1$.

For the harmonic conjugate $ Q $, the ratio changes sign, yielding a division ratio of $ 1 : 4 $ (a harmonized ratio for $-\frac{1}{4} : 1$).

We calculate the coordinates of $ Q $ using the section formula in the new ratio $ 1:4 $:

$ Q = \left( \frac{1 \cdot 6 + 4 \cdot 3}{1+4}, \frac{1 \cdot (-17) + 4 \cdot (-2)}{1+4}, \frac{1 \cdot (-4) + 4 \cdot 2}{1+4} \right) $

Calculating each coordinate:

$ x = \frac{6 + 12}{5} = \frac{18}{5} $

$ y = \frac{-17 - 8}{5} = \frac{-25}{5} = -5 $

$ z = \frac{-4 + 8}{5} = \frac{4}{5} $

Thus, $ Q = \left( \frac{18}{5}, -5, \frac{4}{5} \right) $.

Finally, we find $\alpha + \beta + \gamma$:

$ \alpha + \beta + \gamma = \frac{18}{5} - 5 + \frac{4}{5} = \frac{18}{5} - \frac{25}{5} + \frac{4}{5} = -\frac{3}{5} $

The line $ L $ is the intersection of the two planes given by the equations $ x + 2y + 2z = 15 $ and $ x - y + z = 4 $. The direction ratios of the line $ L $ are determined by the cross product of the normal vectors of these planes.

The normal vector of the first plane is $ \langle 1, 2, 2 \rangle $ and for the second plane is $ \langle 1, -1, 1 \rangle $. We calculate the cross product of these two vectors, represented by the determinant:

$ \left| \begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 2 & 2 \\ 1 & -1 & 1 \end{array} \right| $

This determinant is computed as follows:

$ 4\hat{\mathbf{i}} + \hat{\mathbf{j}} - 3\hat{\mathbf{k}} $

Thus, the direction ratios $ (a, b, c) $ of line $ L $ are $ (4, 1, -3) $.

Finally, we need to find the value of:

$ \frac{a^2 + b^2 + c^2}{b^2} $

Substituting the values, we get:

$ \frac{4^2 + 1^2 + (-3)^2}{1^2} = \frac{16 + 1 + 9}{1} = 26 $

Therefore, the result is 26.

Given that the foot of the perpendicular from $A(1,2,2)$ to the plane $x + 2y + 2z - 5 = 0$ is $B(\alpha, \beta, \gamma)$, we can begin by determining the direction ratios of the perpendicular.

Start by using the formula for the foot of the perpendicular:

$ \frac{\alpha - 1}{1} = \frac{\beta - 2}{2} = \frac{\gamma - 2}{2} = \frac{-(1 + 4 + 4 - 5)}{1 + 4 + 4} $

Simplifying, we get:

$ \Rightarrow \frac{\alpha - 1}{1} = \frac{\beta - 2}{2} = \frac{\gamma - 2}{2} = \frac{-4}{9} $

Solving these equations, we find:

$ \alpha = \frac{5}{9},\ \ \beta = \frac{-8}{9} + 2 = \frac{10}{9},\ \ \gamma = \frac{-8}{9} + 2 = \frac{10}{9} $

Now, calculate $\pi(A)$:

$ \pi(A) = 1 + 2 \times 2 + 2 \times 2 + 5 = 14 $

Next, find $\pi(B)$:

$ \pi(B) = \frac{5}{9} + 2 \left(\frac{10}{9}\right) + 2 \left(\frac{10}{9}\right) + 5 = \frac{5}{9} + \frac{20}{9} + \frac{20}{9} + 5 = 10 $

Finally, compute the ratio $\frac{-\pi(A)}{\pi(B)}$:

$ \frac{-\pi(A)}{\pi(B)} = \frac{-14}{10} = \frac{-7}{5} $

Equation of plane $ \pi_{1} $ is

$ \begin{array}{r} 1(x-3)+2(y+7)-2(z-5)=0 \\\\ x+2 y-2 z-3+14+10=0 \\\\ x+2 y-2 z+21=0 \quad \ldots \text { (i) } \end{array} $

Equation of plane $ \pi_{2} $ is

$ \begin{array}{c} 3(x-2)+2(y-7)+6(z+8)=0 \\\\ 3 x+2 y+6 z-6-14+48=0 \\\\ 3 x+2 y+6 z+28=0\quad \ldots \text { (ii) } \\\\ P_{1}=\frac{21}{\sqrt{1+4+4}}=7 \text { units } \\\\ P_{2}=\frac{28}{\sqrt{9+4+36}}=4 \text { units } \end{array} $

$ P_{1}-P_{2}=(7-4) $ units $ =3 $ units

Given, $ A \equiv(2,3, K), B \equiv(-1, K,-1) $

and $ C \equiv(4,-3,2) $

$ \begin{array}{l} A B=A C \Rightarrow|A B|^{2}=|A C|^{2} \\\\ 9+(K-3)^{2}+(1+K)^{2}=4+36+(K-2)^{2} \\\\ 9+K^{2}+9-6 K+1+K^{2}+2 K \\\\ \qquad 40+K^{2}+4-4 K \\\\ 2 K^{2}-4 K+19-40-K^{2}+4 K-4=0 \\\\ K^{2}=44-19=25 \Rightarrow K=5 \\\\ A \equiv(2,3,5), B \equiv(-1,5,-1), C \equiv(4,-3,2) \\\\ A C=A B=\sqrt{9+4+36}=7 \\\\ B C=\sqrt{25+64+9}=7 \sqrt{2} \end{array} $

Side are $ (7,7,7 \sqrt{2}) $

Equation of plane

$ \begin{array}{l} \left|\begin{array}{ccc} x-1 & y & z+2 \\\\ 3-1 & -1 & 2+2 \\\\ -1 & -3 & 6 \end{array}\right|=0 \\\\ \left|\begin{array}{ccc} x-1 & y & z+2 \\\\ 2 & -1 & 4 \\\\ -1 & -3 & 6 \end{array}\right|=0 \\\\ \Rightarrow(x-1)(-6+12)-y(12+4) \\\\ +(z+2)(-6-1)=0 \\\\ \begin{array}{r} \Rightarrow 6 x-6-16 y-7 z-14=0 \\\\ 6 x-16 y-7 z=20 \end{array} \\\\ \begin{array}{r} x \text {-intercept }=\frac{20}{6}=a \end{array} \\\\ \begin{array}{r} y \text {-intercept }=\frac{-20}{16}=b \end{array} \\\\ z \text {-intercept }=\frac{-20}{7}=c \\\\ 3 a+4 b+7 c=10-5-20=-15 \end{array} $

We have,

$ \begin{aligned} & \mathrm{OA}=\hat{\mathbf{i}}+\hat{\mathbf{j}}, \mathrm{OB}=\hat{\mathbf{j}}+\hat{\mathbf{k}} \\\\ & \mathrm{OC}=\hat{\mathbf{k}}+\hat{\mathbf{i}}, \mathrm{OD}=\hat{\mathbf{i}}-\hat{\mathbf{j}} \text { and } \mathrm{OE}=\hat{\mathbf{j}}-\hat{\mathbf{k}} \\\\ & \therefore \mathrm{CD}=(\hat{\mathbf{i}}-\hat{\mathbf{j}})-(\hat{\mathbf{k}}+\hat{\mathbf{i}})=-\hat{\mathbf{j}}-\hat{\mathbf{k}} \end{aligned} $

$ \begin{aligned} & \mathbf{C E}=(\hat{\mathbf{j}}-\hat{\mathbf{k}})-(\hat{\mathbf{k}}+\hat{\mathbf{i}})=-\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}} \\ & \mathbf{D E}=(\hat{\mathbf{j}}-\hat{\mathbf{k}})-(\hat{\mathbf{i}}-\hat{\mathbf{j}})=-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}} \\ & \mathbf{n}=\mathbf{C D} \times \mathbf{C E}=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 0 & -1 & -1 \\ -1 & 1 & -2 \end{array}\right|=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}} \end{aligned} $

Vector equation of palne,

$ \begin{aligned} & \mathbf{r} \cdot(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})=(\hat{\mathbf{k}}+\hat{\mathbf{i}}) \cdot(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}) \\ \Rightarrow & \mathbf{r} \cdot(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})=2 \quad \ldots \text { (i) } \end{aligned} $

Vector equation line,

$ \begin{aligned} \mathbf{r} & =(\hat{\mathbf{i}}+\hat{\mathbf{j}})+t[(\hat{\mathbf{j}}+\hat{\mathbf{k}})-(\hat{\mathbf{i}}+\hat{\mathbf{j}})] \\ \Rightarrow \mathbf{r} & =(\mathrm{l}-t \hat{\mathbf{i}}+\hat{\mathbf{j}}+t \hat{\mathbf{k}} \quad \ldots \text { (ii) } \end{aligned} $

From Eqs. (i) and (ii), we get

$ \begin{aligned} & {[(1-t) \hat{\mathbf{i}}+\hat{\mathbf{j}}+t \hat{\mathbf{k}}] \cdot(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})=2 } \\ \Rightarrow & 3(1-t)+1-t=2 \Rightarrow 3-3 t+1-t=2 \\ \Rightarrow & \quad-4 t=-2 \Rightarrow t=\frac{1}{2} \end{aligned} $

On putting $t=\frac{1}{2}$ in Eq. (ii), we get

$ \begin{aligned} \mathbf{r} & =\left(1-\frac{1}{2}\right) \hat{\mathbf{i}}+\hat{\mathbf{j}}+\frac{1}{2} \hat{\mathbf{k}} \\ & =\frac{1}{2} \hat{\mathbf{i}}+\hat{\mathbf{j}}+\frac{1}{2} \hat{\mathbf{k}}, \text { which is the required } \end{aligned} $

point of intersection.

Normal vector to the plane

$ x+y-z+4=0 \text { is } \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}} $

Normal vector to the plane

$ 2 x-y+z+1=0 \text { is } 2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}} $

Given planes are perpendicular to the plane ( $\pi$ )

$\therefore$ Normal vector to the plane $(\pi)$ is given by

$ \begin{aligned} \left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\\\ 1 & 1 & -1 \\\\ 2 & -1 & 1 \end{array}\right| & =0 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} \\\\ & =-3(0 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \end{aligned} $

Direction Ratio's of $-3(0 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})$ is ( $0,1,1$ )

Plane $(\pi)$ is passing through the point ( $1,2,-3$ )

$\therefore$ Plane $(\pi) 0(x-1)+1(y-2)+1(z+3)=0$

$ \begin{array}{lr} \Rightarrow & y-2+z+3=0 \\\\ \Rightarrow & 0 x+y+z+1=0 \end{array} $

Given Plane $(\pi) a x+b y+c z+1=0$

$ \begin{aligned} & \therefore \quad a=0, b=1 \text { and } c=1 \\\\ & \text { Now, } a^2+b^2+c^2=0^2+1^2+1^2 \\\\ & =0+1+1=2 \end{aligned} $

We have a variable point $P(x, y, z)$

Distance of $P$ from $X$-axis

$ =\sqrt{0^2+y^2+z^2}=\sqrt{y^2+z^2} $

and distance of $P$ from $Y Z$-plane $=|x|$

Now, $\frac{\sqrt{y^2+z^2}}{|x|}=\frac{2}{3} \quad \text { [given] }$

$ \Rightarrow \frac{y^2+z^2}{x^2}=\frac{4}{9} \Rightarrow 4 x^2-9 y^2-9 z^2=0 $

which is the equation of locus of $P$.

We have,

$ \begin{aligned} & l-m+n=0 \Rightarrow m=n+l\quad \ldots \text { (i) } \\\\ & \text { and } 2 l m-3 m n+n l=0 \\\\ & \Rightarrow 2 l(n+l)-3(n+l) n+n l=0 \end{aligned} $

$ \begin{aligned} & \Rightarrow 2 l^2-3 n^2=0 \quad \text { [from Eq. (i)] } \\\\ & (\sqrt{2} l+\sqrt{3} n)(\sqrt{2} l-\sqrt{3} n)=0 \\\\ & \Rightarrow \quad n=\frac{\sqrt{2}}{\sqrt{3}} l, \frac{-\sqrt{2}}{\sqrt{3}} l \\\\ & \text { Since, } n=\frac{\sqrt{2}}{\sqrt{3}} l \\\\ & \Rightarrow \quad m=\left(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}}\right)l \quad \text { [from Eq. (i)] } \\\\ & \text { and } n=-\frac{\sqrt{2}}{\sqrt{3}} l \Rightarrow m=\left(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}}\right) l \quad \text { [from Eq. (i)] } \end{aligned} $

So, $\frac{l}{1}=\frac{n}{\frac{\sqrt{2}}{\sqrt{3}}}=\frac{m}{\left(\frac{\sqrt{2}+\sqrt{3}}{\sqrt{3}}\right)}$ and

$ \frac{l}{1}=\frac{n}{-\frac{\sqrt{2}}{\sqrt{3}}}=\frac{m}{\left(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}}\right)} $

$ \begin{aligned} & \text { Now, } \\\\ & \begin{aligned} & \cos \theta=\frac{11+\frac{\sqrt{2}}{\sqrt{3}} \cdot\left(-\frac{\sqrt{2}}{\sqrt{3}}\right)+\left(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}}\right)\left(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}}\right)}{\sqrt{(1)^2+\left(\frac{\sqrt{2}}{\sqrt{3}}\right)^2+\left(\frac{\sqrt{2}+\sqrt{3})^2}{\sqrt{3}}\right)^2}} \\\\ &=\frac{1-\frac{2}{3}+\frac{1}{3}}{\left.\sqrt{\frac{5}{3}+\frac{5+2 \sqrt{6}}{3}} \cdot \sqrt{\frac{5}{3}+\frac{5-2 \sqrt{6}}{3}}\right)^2+\left(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}}\right)^2} \\&=\frac{2 / 3}{\sqrt{\frac{2}{3}(5+\sqrt{6})} \cdot \sqrt{\frac{2}{3}(5-\sqrt{6})}} \end{aligned} \\\\ & \Rightarrow \cos \theta=\frac{1}{\sqrt{19}} \end{aligned} $

We have,

$ \begin{aligned} & \text { (c) We have, } \\\\ & A \equiv(5,1,2), B \equiv(3,-4,6), C \equiv(7,0,-1) \end{aligned} $

So, normal vector to plane $\pi$ is

$ \mathbf{n}=\mathbf{A B} \times \mathbf{A C} $

Here, $\mathbf{A B}=-2 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$

$ A C=2 \hat{\mathbf{i}}-1 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} $

So, $\mathbf{n}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\\\ -2 & -5 & 4 \\\\ 2 & -1 & -3\end{array}\right|$

$ \mathbf{n}=19 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+12 \hat{\mathbf{k}} $

Now, equations of plane through point $A$ with normal vector $(19,2,12)$ is

$ 19(x-5)+2(y-1)+12(z-2)=0 $

So, $P=\frac{|-95-2-24|}{\sqrt{361+4+144}}=\frac{121}{\sqrt{509}}$

Direction cosines of $\mathbf{n}$ is $l=\frac{19}{\sqrt{509}}$,

$ m=\frac{2}{\sqrt{509}}, n=\frac{12}{\sqrt{509}} $

Hence,

$ |3 l+2 m+5 n|=\left|\frac{57+4+60}{\sqrt{509}}\right|=\frac{121}{\sqrt{509}}=P $

$A(1,2,3), B(3,-1,5), C(4,0,-3)$

Here,

$ \begin{aligned} & \mathbf{A B}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\ & \mathbf{A C}=3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-6 \hat{\mathbf{k}} \end{aligned} $

$\mathbf{A B} \cdot \mathbf{A C}=6+6-12=0$

$\mathbf{A B} \perp \mathbf{A C} \Rightarrow$ triangle is right angled at $A$. Hence, circumcentre will be mid-point of $B$ and $C$.

Circumcentre $O\left(\frac{3+4}{2}, \frac{-1+0}{2}, \frac{5-3}{2}\right)$

$ \equiv O\left(\frac{7}{2}, \frac{-1}{2}, 1\right) $

$ \Rightarrow|\alpha|+|\beta|=\frac{7}{2}+\frac{1}{2}=4=4|\gamma| $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } l+m+n=0 \Rightarrow l=-(m+n) \\ & \Rightarrow \quad 2 l m+2 n l-m n=0 \\ & \Rightarrow \quad-2(m+n)^2-m n=0 \\ & \Rightarrow \quad 2 m^2+2 n^2+5 m n=0 \Rightarrow \frac{m}{n}=-2,-\frac{1}{2} \\ & \text { At } \frac{m}{n}=-2 \Rightarrow m=-2 n, l=n, n=n \\ & \Rightarrow 1,-2,1 \\ & \text { At } \frac{m}{n}=\frac{-1}{2} \Rightarrow n=-2 m, l=m, m=m \\ & \Rightarrow 1,1,-2 \\ & \cos \theta=\left|\frac{1-2-2}{\sqrt{1+4+1} \sqrt{1+1+4}}\right|=\frac{1}{2} \end{aligned} \end{aligned} $

We have, $a x+b y+c z=2$

Foot of perpendicular is given by

$ \frac{x-1}{a}=\frac{y-0}{b}=\frac{z+2}{c}=\frac{-(a+0-2 c-2)}{a^2+b^2+c^2} $

Given, foot of perpendicular is ( $2,0,-1$ )

$ \begin{aligned} & \Rightarrow \frac{2-1}{a}=\frac{0-0}{b}=\frac{-1+2}{c}=\frac{-(a-2 c-2)}{a^2+b^2+c^2} \\ & \Rightarrow \quad \frac{1}{a}=\frac{0}{b}=\frac{1}{c}=\frac{2 c-a+2}{a^2+b^2+c^2} \\ & \qquad b=0, a=c \\ & \text { and } \frac{1}{a}=\frac{2 a-a+2}{a^2+b^2+c^2} \\ & \Rightarrow \quad \frac{2 a^2}{a}=a+2 \Rightarrow a=2 \Rightarrow c=2 \\ & \therefore a^2+b^2+c^2=4+0+4=8 \end{aligned} $

Given as, $A(1,2,3), B(3,7,-2), C(6,7,7)$ and $D(-1,0,-1)$

Centroid of $\triangle A B D=(1,3,0) \rightarrow \hat{\mathbf{i}}+3 \hat{\mathbf{j}}$

Centroid of $\triangle A C D=(2,3,3)$

$ \begin{aligned} \rightarrow & 2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\ \therefore \mathbf{r} & =\mathbf{a}+t(\mathbf{b}-\mathbf{a}) \\ \mathbf{r} & =(\hat{\mathbf{i}}+3 \hat{\mathbf{j}})+t(\hat{\mathbf{i}}+3 \hat{\mathbf{k}}) \\ \therefore \mathbf{r} & =(1+t) \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \end{aligned} $

Let $A\left(x_1, y_1, z_1\right), B\left(x_2, y_2, z_2\right)$ and $C\left(x_3, y_3, z_3\right)$ are vertices of a triangle.

∵ Mid-point of $A B=(3,0,0)$

$ \therefore x_1+x_2=6, y_1+y_2=0, z_1+z_2=0 $

Similarly, $(0,4,0)$ and $(0,0,5)$ are mid-points of $B C$ and $C A$.

$ \begin{aligned} & \therefore \quad x_2+x_3=0, y_2+y_3=8, z_2+z_3=0 \\ & \text { and } x_1+x_3=0, y_1+y_3=0, z_1+z_3=10 \end{aligned} $

By solving these equations, we get

$ \begin{aligned} & x_1=3, y_1=-4, z_1=5, x_2=3, y_2=4 \\ & z_2=-5, x_3=-3, y_3=4, z_3=5 \end{aligned} $

So, $A(3,-4,5), B(3,4,-5)$ and $C(-3,4,5)$

$ \begin{aligned} & A B=\sqrt{0+(8)^2+(10)^2}=\sqrt{164} \\ & B C=\sqrt{(6)^2+0+(10)^2}=\sqrt{136} \end{aligned} $

and $C A=\sqrt{(-3-3)^2+(4+4)^2+0}=\sqrt{100}$

$ A B^2+B C^2+C A^2=164+136+100=400 $

Let $\mathbf{x}$ and $\mathbf{y}$ be two vectors whose direction cosines are $l, m, n$ and $a, b, c$ respectively.

$ \begin{aligned} & \therefore \quad \hat{\mathbf{x}}=l \hat{\mathbf{i}}+m \hat{\mathbf{j}}+n \hat{\mathbf{k}} \\ & \text { and } \hat{\mathbf{y}}=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}=c \hat{\mathbf{k}} \end{aligned} $

Now, $\cos \theta=\hat{\mathbf{x}} \cdot \hat{\mathbf{y}}$

$ \Rightarrow \quad \hat{\mathbf{x}} \cdot \hat{\mathbf{y}}=l a+m b+n c $

For perpendicular $\hat{\mathbf{x}} \cdot \hat{\mathbf{y}}=0$

$ \therefore \quad l a+m b+n c=0 $

For parallel lines

$ \begin{array}{rlrl} & \hat{\mathbf{x}} =\lambda \hat{\mathbf{y}} \\ & \therefore l=\lambda a, \ m & =\lambda b \text { and } n=\lambda c \\ \Rightarrow & \frac{l}{a} =\frac{m}{b}=\frac{n}{c} \end{array} $

and the direction ratios of the lines bisecting the angle between is $l \pm a$, $m \pm b, n \pm c$.

Let $P(2,-1,3)$ be the foot of the perpendicular from the origin to the plane

$ \therefore \mathbf{O P}=\hat{\mathbf{n}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}} $

and

$ \begin{aligned} & p=|2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}| \\ & =\sqrt{r^2+{1^2}^2+3^2}=\sqrt{4+1+9}=\sqrt{14} \end{aligned} $

$ \therefore \hat{\mathbf{n}}=\frac{2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}}{\sqrt{14}} $

The vector equation of the plane is

$ \begin{aligned} & \mathbf{r} \cdot \hat{\mathbf{n}}=p \\ & (x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \cdot \frac{2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}}{\sqrt{14}}=\sqrt{14} \\ & (x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \cdot(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}})=14 \\ & 2 x-y+3 z=14 \\ & 2 x-y+3 z-14=0 \end{aligned} $

Given, $P$ divides $A B$ in the ratio $1: 2$.

∴ Coordinates of

$ P=\left(\frac{1 \times 2+2 \times 1}{1+2}, \frac{3 \times 1+2 \times 2}{1+2}, \frac{1 \times 1+3 \times 2}{1+2}\right) $

(by section formula)

$ \equiv\left(\frac{4}{3}, \frac{7}{3}, \frac{7}{3}\right) $

Coordinates of $Q$ divides $B C$ in the ratio -2:3.

∴ Coordinates of $Q$

$ \begin{aligned} & =\left(\frac{3 \times 2-2 \times 3}{1}, \frac{3 \times 3-2 \times 1}{1}, \frac{3 \times 1-4}{1}\right) \\ & =(0,7,-1) \end{aligned} $

Distance of $P Q$

$ \begin{aligned} & =\sqrt{\left(0-\frac{4}{3}\right)^2+\left(7-\frac{7}{3}\right)^2+\left(-1-\frac{7}{3}\right)^2} \\ & =\sqrt{\frac{16}{9}+\frac{196}{9}+\frac{100}{9}} \\ & =\sqrt{\frac{312}{9}}=\frac{2 \sqrt{78}}{3} \end{aligned} $

$ \begin{aligned} &\text { Equation of the line } B C\\ &\Rightarrow \quad \frac{x-2}{1}=\frac{y-2}{1}=\frac{z-1}{-1} \end{aligned} $

∵ Coordinates of point $M$

$ \equiv(\lambda+2, \lambda+2,-\lambda+1) $

Direction ratio of $A M$

$ \begin{array}{r} =(\lambda+2-1, \lambda+2+2,-\lambda+1-1) \\ =(\lambda+1, \lambda+4,-\lambda) \end{array} $

Now, $A M$ is perpendicular to $B C$

$ \begin{aligned} \therefore \lambda+1+\lambda+4+\lambda & =0 \\ 3 \lambda+5 & =0 \\ \Rightarrow \quad \lambda & =\frac{-5}{3} \end{aligned} $

∴ Coordinate of $M$

$ =\left(\frac{-5}{3}+2, \frac{-5}{3}+2, \frac{5}{3}+1\right)=\left(\frac{1}{3}, \frac{1}{3}, \frac{8}{3}\right) $

Now, coordinates of

$ \begin{aligned} & A^{\prime}=\left(\frac{2}{3}-1, \frac{2}{3}+2, \frac{16}{3}-1\right) \\ & \quad(l, m, n) \equiv\left(\frac{-1}{3}, \frac{8}{3}, \frac{13}{3}\right) \\ & l^2+m^2+n^2=\frac{1}{9}+\frac{64}{9}+\frac{169}{9}=\frac{234}{9}=26 \end{aligned} $

Let $A(1,1,1)$ is perpendicular to the line joining the points $P(6,3,2)$ and $Q(1,-4,-9)$

We know that direction ratios of a line ( $l$ )

$ \equiv\left(x_2-x_1, y_2-y_1, z_2-z_1\right) $

DR's of a line $\equiv(1-6,-4-3,-9-2)$

$ \equiv(-5,-7,-11) $

DR's of a normal (n) = DR's of a line

$ =(-5,-7,-11) $

Let $\left(a_1, b_1, c_1\right)=(-5,-7,-11)$

and $A(1,1,1) \equiv A\left(x_1, y_1, z_1\right)$

Equation of the plane is

$ \begin{aligned} & & a_1\left(x-x_1\right)+b_1\left(y-y_1\right)+c_1\left(z-z_1\right) & =0 \\ \Rightarrow & & -5(x-1)-7(y-1)-11(z-1) & =0 \\ \Rightarrow & & 5(x-1)+7(y-1)+11(z-1) & =0 \\ \Rightarrow & & 5 x+7 y+11 z-23 & =0 \end{aligned} $

On comparing the above equation with $a x+b y+c z-23=0$, we get

$ a=5, b=7 \text { and } c=11 $

Now, $a+b-c=5+7-11=1$

Equation of line passing through

$ \begin{aligned} & \hat{\mathbf{i}}-\hat{\mathbf{j}+0 \mathbf{k}, 0 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}} \\ & \frac{x-1}{0-1}=\frac{y+1}{1+1}=\frac{z-0}{-1-0} \\ & \frac{x-1}{-1}=\frac{y+1}{2}=\frac{z}{-1}=r\,\,\,\,\,\, (let)...(i)\end{aligned} $

Equation of plane passing through

$ \begin{aligned} & 2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+0 \hat{\mathbf{k}}, 0 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}, \hat{\mathbf{i}}+0 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\ & \qquad a(x-2)+b(y-1)+c(z-0)=0 \\ & \Rightarrow \quad-2 a+b-c=0 \\ & \text { and } \quad-a-b+2 c=0 \\ & \Rightarrow \quad 2 a-b+c=0 \\ & \text { and } \quad a+b-2 c=0 \end{aligned} $

$ \begin{aligned} a+b-2 c & =0 \\ \frac{a}{2-1} & =\frac{b}{1+4}=\frac{c}{2+1} \\ \frac{a}{1} & =\frac{b}{5}=\frac{c}{3} \end{aligned} $

$ \begin{aligned} & \therefore \quad(x-2)+5(y-1)+3(z-0)=0 \\ & \Rightarrow \quad x+5 y+3 z=7\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{aligned} $

Point on line (i) is

$ \begin{aligned} &(1-r, 2 r-1,-r) \text { lies on plane (ii) } \\ & 1-r+5(2 r-1)+3(-r)=7 \\ & 1-r+10 r-5-3 r=7 \\ & \Rightarrow \quad r=11 / 6 \end{aligned} $

Point of intersection

$ \left(-\frac{5}{6}, \frac{16}{6},-\frac{11}{6}\right) $

or $\quad \frac{1}{6}(-5 \hat{\mathbf{i}}+16 \hat{\mathbf{j}}-11 \hat{\mathbf{k}})$

∵ Let the point $P(x, y, z)$ in the plane $\pi$.

Now, If plane $\pi$ is parallel to the plane having normal vector is $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$.

So, plane $\pi$ having normal vector as $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$

∴ Direction cosines of palne $\pi$ is $(x-3) \hat{\mathbf{i}}(y-4) \hat{\mathbf{j}}+(z-5) \hat{\mathbf{k}}$ with point $3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+5 \hat{\mathbf{k}}$.

So, $(x-3) \cdot 1+(y+4) 2+(z-5)(-3)=0$

$ \begin{array}{lr} \Rightarrow & x-3+2 y+8-3 z+15=0 \\ \therefore & x+2 y-3 z+20=0 \end{array} $

Given, lines

$ 3 l+2 m+n=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and  $ 2 m n-3 n l+5 l m=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

From Eq. (i), we get

$ n=-(3 l+2 m) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)$

On putting in Eq. (ii), we get

$ \begin{array}{lr} & -2 m(3 l+2 m)+3 l(3 l+2 m)+5 l m=0 \\ \Rightarrow & -6 l m-4 m^2+9 l^2+6 l m+5 l m=0 \\ \Rightarrow & 9 l^2+5 l m-4 m^2=0 \\ \Rightarrow & 9 l^2+9 l m-4 l m-4 m^2=0 \\ \Rightarrow & 9 l(l+m)-4 m(l+m)=0 \\ \Rightarrow & (l+m)(9 l-4 m)=0 \\ \Rightarrow & l=-m \text { or } l=\frac{4 m}{9} \end{array} $

When $l=-m$, then from Eq. (iii), we get

$ \begin{aligned} & & n & =-(-3 m+2 m)=m \\ \Rightarrow & & -l & =m=n \\ \Rightarrow & & \frac{l}{-1} & =\frac{m}{1}=\frac{n}{1} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\end{aligned} $

When $l=\frac{4}{9} m$.

$ \begin{aligned} n & =-\left(\frac{4}{3} m+2 m\right)=-\frac{10}{3} m \\ \Rightarrow \quad \frac{l}{4 / 9} & =m=\frac{n}{-10 / 3} \\ \Rightarrow \quad \frac{l}{4} & =\frac{m}{9}=\frac{n}{-30} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)\end{aligned} $

Angle between lines whose $\mathrm{DC}^{\prime}$ 's are proportional given in Eqs. (iv) and (v),

$ \cos \theta=\frac{(-1)(4)+(1)(9)+1(-30)}{\sqrt{1+1+1} \sqrt{16+81+900}}=\frac{-25}{\sqrt{2991}} $