Three Dimensional Geometry

Equation of plane parallel to

$ \begin{aligned} & x-y-z=0 \text { is } \\ & x-y-z+\lambda=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

Plane Eq. (i) passes through ( $1,-2,1$ ).

$ \begin{aligned} \therefore & & 1+2-1+\lambda & =0 \\ \Rightarrow & & \lambda & =-2 \end{aligned} $

Equation of plane

$ x-y-z-2=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

On comparing with

$ a x+b y+c z-2=0 $

We have, $\quad a=1, b=-1, c=-1$

$ \begin{aligned} \therefore \quad b-2 c & =-1-2(-1) \\ & =-1+2=1=a \end{aligned} $

Now, first to find equation of plane passing through point $A(3,-2,1)$ and perpendicular to the vector $4 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$.

$ \begin{aligned} & \text { So, } \mathbf{n} \cdot(4 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-4 \hat{\mathbf{k}})=(4 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}) \\ & \cdot(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}) \\ & \begin{aligned} & \Rightarrow(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \cdot(4 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}) \\ &=12-14-4=-6 \\ & 4 x+7 y-4 z=-6 \\ & \Rightarrow 4 x+7 y-4 z+6=0 \end{aligned} \end{aligned} $

Since, $M$ is the foot of the perpendicular drawn from $P(1,2,-1)$ to the plane $4 x+7 y-4 z+6=0$

So,

$ \begin{aligned} P M & =\left|\frac{4(1)+7(2)-4(-1)+6}{\sqrt{16+49+16}}\right| \\ & =\left|\frac{4+14+4+6}{\sqrt{81}}\right|=28 / 9 \end{aligned} $

Given, $A=(1,-1,2), B=(3,4,-2)$, $C=(0,3,2)$ and $D=(3,5,6)$

Now, DR's of the lines $\mathbf{A B}$ is $(2,5,-4)$ and DR's of the lines $\mathbf{C D}$ is $(3,2,4)$ Let ' $\theta$ ' be the angle between them, we get

$ \begin{aligned} \cos \theta & =\frac{\mathbf{A B} \cdot \mathbf{C D}}{|\mathbf{A B}| \cdot|\mathbf{C D}|} \\ & =\frac{6+10-16}{\sqrt{2^2+5^2+4^2} \sqrt{3^2+2^2+4^2}} \\ \Rightarrow \cos \theta & =0 \\ \therefore \quad \theta & =90^{\circ} \end{aligned} $

Assertion (A) DR'S of a line $L_1$ is 2, 5, 7 and

DR'S of a line $L_2$ is $\frac{4}{\sqrt{19}}, \frac{10}{\sqrt{19}}, \frac{14}{\sqrt{19}}$

Thus, $\frac{2}{\frac{4}{\sqrt{19}}}=\frac{5}{\frac{10}{\sqrt{19}}}=\frac{7}{\frac{14}{\sqrt{19}}}=\frac{1}{2}$

Hence, the lines $L_1$ and $L_2$ are parallel. Reason (R) If the DR'S of a line $L_1$ are $a_1, b_1, c_1$ and the $\mathrm{DR}^{\dagger} \mathrm{S}$ of a line $L_2$ are $a_2, b_2, c_2$ and $a_1 a_2+b_1 b_2+c_1 c_2=0$, then the lines are not parallel.

So, Assertion (A) is true but Reason (R) is false.

The given planes are $2 x+3 y+z=1$ and $x+3 y+2 z=2$

$ \begin{aligned} & \text { Thus, } \hat{\mathbf{n}}=\left|\begin{array}{lll} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 3 & 1 \\ 1 & 3 & 2 \end{array}\right|=\hat{\mathbf{i}}(3)+\hat{\mathbf{j}}(-3)+\hat{\mathbf{k}}(3) \\ & =3 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\ & \text { and } \quad \hat{\mathbf{a}}=\frac{\mathbf{n}}{|\mathbf{n}|}=\frac{3 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}}{\sqrt{9+9+9}}=\frac{\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}}{\sqrt{3}} \\ & \Rightarrow \quad \mathbf{a}=\frac{1}{\sqrt{3}} \hat{\mathbf{i}}-\frac{1}{\sqrt{3}} \hat{\mathbf{j}}+\frac{1}{\sqrt{3}} \hat{\mathbf{k}} \end{aligned} $

Since, the line ' $L$ ' makes an angle $\alpha$ with the positive direction of $X$-axis, then

$ \cos \alpha=\frac{1}{\sqrt{3}} $

Given the direction cosines $(l, m, n)$ of two lines, we have the following relationships:

$l + m + n = 0$

$lm = 0$

From the equation $lm = 0$, we infer that either $l = 0$ or $m = 0$.

Case 1: $l = 0$

Substitute $l = 0$ into the first equation: $m + n = 0$.

This gives $m = -n$.

Now, using the condition for direction cosines, $l^2 + m^2 + n^2 = 1$, we substitute $m = -n$ and $l = 0$:

$ 0^2 + (-n)^2 + n^2 = 1 \implies 2n^2 = 1 \implies n = \frac{1}{\sqrt{2}} $

Therefore, the set of direction cosines is:

$ \langle l, m, n \rangle = \left\langle 0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right\rangle $

Case 2: $m = 0$

Substitute $m = 0$ into the first equation: $l + n = 0$.

This gives $l = -n$.

Again using $l^2 + m^2 + n^2 = 1$ with $l = -n$ and $m = 0$:

$ (-n)^2 + 0^2 + n^2 = 1 \implies 2n^2 = 1 \implies n = \frac{1}{\sqrt{2}} $

Thus, the set of direction cosines is:

$ \langle l, m, n \rangle = \left\langle -\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}} \right\rangle $

Finding the Angle $\theta$ Between the Lines

The cosine of the angle $\theta$ between the two lines can be calculated using the dot product of their direction cosine vectors. Let us compute it:

$ \cos \theta = 0 \times \left(-\frac{1}{\sqrt{2}}\right) + \left(-\frac{1}{\sqrt{2}}\right) \times 0 + \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = \frac{1}{2} $

Thus, the angle $\theta$ is:

$ \theta = \frac{\pi}{3} $

Perpendicular distance on $X$-axis

$ =y^2+z^2 $

Perpendicular distance on $Y$-axis

$ =x^2+z^2 $

Perpendicular distance on Z-axis

$ \begin{gathered} =x^2+y^2 \\ \therefore 2\left(x^2+y^2+z^2\right)=k\left(x^2+y^2+z^2\right) \end{gathered} $

Hence, $2=k$

Plane is,

$ \overline{\mathbf{r}} \cdot \overline{\mathbf{n}}=\overline{\mathbf{a}} \cdot \overline{\mathbf{n}} $

$ \begin{aligned} & \overline{\mathbf{r}} \cdot(-5 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+11 \hat{\mathbf{k}})=-\frac{15}{2}-\frac{21}{2}-\frac{99}{2} \\ \Rightarrow \quad & \quad-5 x-3 y+11 z=-\frac{135}{2} \\ \Rightarrow & -10 x-6 y+22 z+135=0 \\ \Rightarrow \quad & 10 x+6 y-22 z-135=0 \end{aligned} $

$ \text { } \begin{aligned} & \mathbf{r}=(2-\lambda+\mu) \hat{\mathbf{i}}+(1-\mu) \hat{\mathbf{j}} +(2-3 \lambda+2 \mu) \hat{\mathbf{k}} \\ & \mathbf{r}=(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}})+\lambda(-\hat{\mathbf{i}}-3 \hat{\mathbf{k}}) +\mu(\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \end{aligned} $

which is a plane through $\mathbf{a}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and parallel to the vectors.

$ \begin{aligned} & \mathbf{b}=-\hat{\mathbf{i}}-3 \hat{\mathbf{k}} \\ \text { and } & \mathbf{c}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}} \end{aligned} $

Therefore, it is perpendicular to the vector.

$ \mathbf{n}=\mathbf{b} \times \mathbf{c} $

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

Hence, its vector equation is

$ \begin{aligned} & (\mathbf{r}-\mathbf{a}) \mathbf{n}=\mathbf{0} \\ \Rightarrow & \mathbf{r} \cdot \mathbf{n}=\mathbf{a} \cdot \mathbf{n} \\ \Rightarrow & \mathbf{r} \cdot(-3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})=2(-3)+1(-1)+2(1) \\ & \mathbf{r} \cdot(-3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})=-6-1+2 \\ & \mathbf{r} \cdot(-3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})=-5 \end{aligned} $

Cartesian equation,

$ \begin{aligned} & (x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \cdot(-3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})=-5 \\ \Rightarrow \quad & -3 x-y+z=-5 \\ \Rightarrow \quad & 3 x+y-z=5 \end{aligned} $

Option (a) is correct.

Equation of the passing through points

$(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}})$ and $(-\hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}})$ is

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

Vector perpendicular to plane

$ \pi_1:(-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) $

Vector perpendicular to plane

$ \pi_2:(4 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}) $

We know angle between two planes is same as angle between vectors perpendicular to them. So, let angle be $\theta$, then

$ \begin{aligned} \cos \theta & =\frac{(-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \cdot(4 \hat{\mathbf{i}}-5 \hat{\mathbf{j}})}{|-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}||4 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}|} \\ & =\frac{5}{\sqrt{5} \sqrt{41}}=\frac{\sqrt{5}}{\sqrt{41}}=\sqrt{\frac{5}{41}} \end{aligned} $

Given, $A=(1,2,0), B=(2,0,-1) C=(0,-2,3)$ and $D=(-1,2,-3)$. $G_1$ is centroid of $\triangle A B C$.

$ \begin{aligned} \therefore G_1 & =\left(\frac{1+2+0}{3}, \frac{2+0-2}{3}, \frac{0-1+3}{3}\right) \\ & =\left(1,0, \frac{2}{3}\right) \end{aligned} $

$G_2$ is centroid of tetrahedron $A B C D$.

$ G_2=\left(\frac{1+2+0-1}{4}, \frac{2+0-2+2}{4},\frac{0-1+3-3}{4}\right) $

$ =\left(\frac{1}{2}, \frac{1}{2}, \frac{-1}{4}\right) $

$ P(x, y, z)=\left(\frac{4 \times \frac{1}{2}+3 \times 1}{4+3}, \frac{4 \times \frac{1}{2}+3 \times 0}{4+3}\begin{aligned} , \left.\frac{4 \times\left(-\frac{1}{4}\right)+3\left(\frac{2}{3}\right)}{4+3}\right) \end{aligned}\right.$

$ \begin{aligned} & =\left(\frac{2+3}{7}, \frac{2+0}{7}, \frac{-1+2}{7}\right) \\ & =\left(\frac{5}{7}, \frac{2}{7}, \frac{1}{7}\right) \end{aligned} $

Given, $a-b+c=0$

$ \begin{array}{lr} \Rightarrow & b=a+c \,\,...(i)\\ \text { and } & a^2-b^2+2 c^2=0 \\ \Rightarrow & a^2-(a+c)^2+2 c^2=0 \end{array} $

[using Eq. (i)]

$ \begin{array}{lr} \Rightarrow & a^2-a^2-c^2-2 a c+2 c^2=0 \\ \Rightarrow & c^2-2 a c=0 \\ \Rightarrow & c(c-2 a)=0 \\ \Rightarrow & c=0 \text { or } c=2 a \end{array} $

When $c=0$, then $b=a$ and when $c=2 a$, then $b=3 a$

∴ Direction ratios of Eq. (ii) are $(1,1,0)$ and direction ratios of (iii) are ( $1,3,2$ ).

$ \begin{aligned} \therefore \cos \theta & =\frac{1 \times 1+1 \times 3+0 \times 2}{\sqrt{1^1+1^2+0} \sqrt{1^2+3^2+2^2}} \\ & =\frac{4}{\sqrt{2} \sqrt{14}}=\frac{2}{\sqrt{7}} \end{aligned} $

Let the given three point be $A \equiv(0,1,2), B \equiv(3,0,2)$ and $C \equiv(4,5,0)$ Equation of plane passing through three points is.

$ \begin{array}{rlrl} & \left|\begin{array}{ccc} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{array}\right| & =0 \\ \Rightarrow & \left|\begin{array}{ccc} x-0 & y-1 & z-2 \\ 3-0 & 0-1 & 2-2 \\ 4-0 & 5-1 & 0-2 \end{array}\right| =0 \\ \Rightarrow & \left|\begin{array}{ccc} x & y-1 & z-2 \\ 3 & -1 & 0 \\ 4 & 4 & -2 \end{array}\right| =0 \end{array} $

$ \begin{array}{rlrl} & \Rightarrow & (z-2)[(3)(4)-(4)(-1)]-0 +(-2)[(x)(-1)-3(y-1)] & =0 \\ & \Rightarrow & (z-2)[12+4]-2[-x-3 y+3] & =0 \\ & \Rightarrow & 16 z-32+2 x+6 y-6 & =0 \\ & \Rightarrow & 2 x+6 y+16 z & =38 \\ & \Rightarrow & x+3 y+8 z & =19 \end{array} $

$ \begin{aligned} &\begin{aligned} \Rightarrow & \frac{x+3 y+8 z}{\sqrt{1+9+64}}=\frac{19}{\sqrt{1+9+64}} \\ & \frac{1}{\sqrt{74}} x+\frac{3}{\sqrt{74}} y+\frac{8}{\sqrt{74}} z=\frac{19}{\sqrt{74}} \end{aligned}\\ &\text { Comparing to } l x+m y+n z=p \text {, we have }\\ &\begin{array}{ll} & l=\frac{1}{\sqrt{74}}, m=\frac{3}{\sqrt{74}} \text { and } n=\frac{8}{\sqrt{74}} \\ \therefore & |l|+|m|+|n|=\frac{1+3+8}{\sqrt{74}}=\frac{12}{\sqrt{74}} \end{array} \end{aligned} $

Equation of line $L$ passing through points

$ \begin{aligned} & 2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+8 \hat{\mathbf{k}} \text { and } \hat{\mathbf{i}}+6 \hat{\mathbf{j}}+4 \hat{\mathbf{k}} \text { is } \\ & \mathbf{r}=(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+8 \hat{\mathbf{k}})+\lambda[(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+8 \hat{\mathbf{k}}) \\ & \qquad \begin{array}{r} \quad-(\hat{\mathbf{i}}+6 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})] \\ L: \mathbf{r}=(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+8 \hat{\mathbf{k}})+\lambda(\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \end{array} \\ & L: \frac{x-2}{1}=\frac{y-3}{-3}=\frac{z-8}{4}=\lambda \end{aligned} $

Now, plane is parallel to the vectors $\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$

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

So equation of plane passing through

$ -5 \hat{\mathbf{i}}+19 \hat{\mathbf{j}}-14 \hat{\mathbf{k}} $

$ \begin{aligned} a\left(x-x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right) & =0 \\ \Rightarrow-1(x+5)-2(y-19)-1(z+14) & =0 \\ P: x+2 y+z & =19 \end{aligned} $

Any general point on the line is

$ x=\lambda+2, y=-3 \lambda+3, z=4 \lambda+8 $

So, $A(\lambda+2,-3 \lambda+3,4 \lambda+8)$

Now, this point lies on the plane

$ \begin{gathered} \Rightarrow(\lambda+2)+2(-3 \lambda+3)+(4 \lambda+8)=19 \\ -\lambda=3 \Rightarrow \lambda=-3 \end{gathered} $

So, point of intersection

$ \begin{aligned} & A(\lambda+2,-3 \lambda+3,4 \lambda+8) \\ & A(-3+2,9+3,-12+8) \equiv(-1,12,-4) \end{aligned} $

So position vector of $A$ is $-\hat{\mathbf{i}}+12 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$.

Given, $P_1: \mathbf{r} \cdot(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})=5$

$ P_2: \mathbf{r} \cdot(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})=7 $

Let $\quad \mathbf{n}_1=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$

$ \mathbf{n}_2=\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}} $

Now, $\mathbf{n}=\mathbf{n}_1 \times \mathbf{n}_2$

$ \begin{aligned} \mathbf{n} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 3 & 4 \\ 1 & 1 & -1 \end{array}\right| \\ & =-7 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-\hat{\mathbf{k}} \end{aligned} $

Equation of line of intersection of planes passing through the point $(16,-9,0)$ is

$ \begin{aligned} \mathbf{r} & =(16 \hat{\mathbf{i}}-9 \hat{\mathbf{j}}+0 \hat{\mathbf{k}})+\lambda(-7 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \\ & =(16-7 \lambda) \hat{\mathbf{i}}+(6 \lambda-9) \hat{\mathbf{j}}-\lambda \hat{\mathbf{k}} \end{aligned} $

Let $A(1,1,1), B(1,-4,3), C(2,-2,0)$ and $D(8,1,4)$ are the vertices of a tetrahedron.

Since, $G_1$ is the centroid of face $A B C$

$ \begin{aligned} \therefore G_1 & \equiv\binom{\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3},}{\frac{z_1+z_2+z_3}{3}} \\ G_1 & =\left(\frac{1+1+2}{3}, \frac{1-4-2}{3}, \frac{1+3+0}{3}\right) \\ G_1 & =\left(\frac{4}{3}, \frac{-5}{3}, \frac{4}{3}\right) \end{aligned} $

Now, $G_2$ is the centroid of face $B C D$,

$ \begin{aligned} \therefore G_2 & =\left(\frac{1+2+8}{3}, \frac{-4-2+1}{3}, \frac{3+0+4}{3}\right) \\ & G_2=\left(\frac{11}{3}, \frac{-5}{3}, \frac{7}{3}\right) \end{aligned} $

Also, $G_3$ is the centroid of the face $C D A$.

$ \begin{aligned} \therefore G_3 & =\left(\frac{1+2+8}{3}, \frac{1-2+1}{3}, \frac{1+0+4}{3}\right) \\ G_3 & \equiv\left(\frac{11}{3}, 0, \frac{5}{3}\right) \end{aligned} $

Also, $G_4$ is the centroid of the face $D A B$,

$ \begin{aligned} \therefore G_4 & =\left(\frac{1+1+8}{3}, \frac{1-4+1}{3}, \frac{1+3+4}{3}\right) \\ & G_4=\left(\frac{10}{3}, \frac{-2}{3}, \frac{8}{3}\right) \end{aligned} $

Now, centroid of tetrahedron having $G_1, G_2, G_3$ and $G_4$ as its vertices.

$ G=\left(\frac{\frac{4}{3}+\frac{11}{3}+\frac{11}{3}+\frac{10}{3}}{4}, \frac{-\frac{5}{3}-\frac{5}{3}+0-\frac{2}{3}}{4},\frac{\frac{4}{3}+\frac{7}{3}+\frac{5}{3}+\frac{8}{3}}{4}\right) $

$ \therefore G=(3,-1,2) $

$ \begin{aligned}Dr's\,\, of\,\,\, \mathbf{A B} & =\mathbf{B}-\mathbf{A} \\ & =(4 \hat{\mathbf{i}}+\hat{\mathbf{j}}+0 \hat{\mathbf{k}})-(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \\ & =2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}} \end{aligned} $

$ \begin{aligned}Dr's \,\,of\,\, \mathbf{A} \mathbf{A C} & =\mathbf{C}-\mathbf{A} \\ & =(-\hat{\mathbf{i}}-\hat{\mathbf{j}}+11 \hat{\mathbf{k}})-(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}}) \\ & =-3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+12 \hat{\mathbf{k}} \end{aligned} $

If AD is the angle bisector of $\angle B A C$,

                   $ \begin{aligned} \mathbf{A D} & =\frac{\mathbf{A B}}{|\mathbf{A B}|}+\frac{\mathbf{A C}}{|\mathbf{A C}|} \\ & =\frac{2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}}{3}+\frac{-3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+12 \hat{\mathbf{k}}}{13} \end{aligned} $

                          $ \begin{aligned} &=\frac{17 \hat{\mathbf{i}}-38 \hat{\mathbf{j}}+49 \hat{\mathbf{k}}}{3 \times 13}\\ &\text { Dr's : } \frac{17}{39}, \frac{-38}{39}, \frac{49}{39} \equiv(17,-38,49) \end{aligned} $

A plane passes through

$ A(2,3,0), B(0,-5,2) \text { and } C(-2,0,3) . $

Equation of plane passing through

$ \begin{aligned} & 3 \text { points }\left(x_1, y_1, z_1\right),\left(x_2, y_2, z_2\right)\left(x_3, y_3, z_3\right) \\ & \Rightarrow\left|\begin{array}{ccc} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{array}\right|=0 \\ & \Rightarrow \quad\left|\begin{array}{ccc} x-2 & y-3 & z-0 \\ 0-2 & -5-3 & 2-0 \\ -2-2 & 0-3 & 3-0 \end{array}\right|=0 \\ & \Rightarrow \quad\left|\begin{array}{ccc} x-2 & y-3 & z \\ -2 & -8 & 2 \\ -4 & -3 & 3 \end{array}\right|=0 \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \\ & \Rightarrow \\ & \Rightarrow \end{aligned} \begin{aligned} &-18(x-2)-(y-3)(2)+z(-26)=0 \\ &\,\,\,\,\,\,9(x-2)+(y-3)+13 z=0 \\ & 9 x+y+13 z=21 \end{aligned}\\ &\text { If plane } P \text { cuts } X \text {-axis at } A \text {, then } A(x, 0,0)\\ &\begin{aligned} & \Rightarrow 9 x+0+0=21 \\ & \Rightarrow \quad x=7 / 3 \\ & \text { Point } A \equiv\left(\frac{7}{3}, 0,0\right) \end{aligned} \end{aligned} $

Given, the plane passing through the point

$ \begin{aligned} & \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}, 3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}},-3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}} \\ & (\mathbf{r}-\mathbf{a})[(\mathbf{b}-\mathbf{a}) \times(\mathbf{c}-\mathbf{a})]=0 \\ & \Rightarrow \mathbf{b}-\mathbf{a}=(3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}})-(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}) \\ & =2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+7 \hat{\mathbf{k}} \\ & \Rightarrow \mathbf{c}-\mathbf{a}=(-3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-5 \hat{\mathbf{k}})-(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}) \\ & =-4 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} \end{aligned} $

$ \begin{aligned} & \therefore(\mathbf{b}-\mathbf{a}) \times(\mathbf{c}-\mathbf{a})=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 1 & 7 \\ -4 & 4 & -2 \end{array}\right| \\ & =\hat{\mathbf{i}}(-2-28)-\hat{\mathbf{j}}(-4+28)+\hat{\mathbf{k}}(8+4) \\ & =-30 \hat{\mathbf{i}}-24 \hat{\mathbf{j}}+12 \hat{\mathbf{k}} \\ & \text { Now, }(\mathbf{r}-\mathbf{a}) \cdot[(\mathbf{b}-\mathbf{a}) \times(\mathbf{c}-\mathbf{a})]=0 \\ & \Rightarrow[\mathbf{r}-(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}})][-30 \hat{\mathbf{i}}-24 \hat{\mathbf{j}}+12 \hat{\mathbf{k}}]=0 \\ & \Rightarrow \mathbf{r}(-30 \hat{\mathbf{i}}-24 \hat{\mathbf{j}}+12 \hat{\mathbf{k}}) +30-48+36=0 \end{aligned} $

$ \Rightarrow \mathbf{r}(-30 \hat{\mathbf{i}}-24 \hat{\mathbf{j}}+12 \hat{\mathbf{k}})=-18 $

From the options,

$ \begin{aligned}(a) & -\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} ;(-\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \\ & (-30 \hat{\mathbf{i}}-24 \hat{\mathbf{j}}+12 \hat{\mathbf{k}}) \\ & =30-72-24 \neq-18 \end{aligned} $

$ \begin{aligned}(b) & 7 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}-6 \hat{\mathbf{k}} ;(7 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}) \\ & (-30 \hat{\mathbf{i}}-24 \hat{\mathbf{j}}+12 \hat{\mathbf{k}}) \\ & =-210+120-72 \neq-18 \end{aligned} $

$ \begin{aligned}(c) & -\hat{\mathbf{i}}+9 \hat{\mathbf{j}}+14 \hat{\mathbf{k}} ;(-\hat{\mathbf{i}}+9 \hat{\mathbf{j}}+14 \hat{\mathbf{k}}) \\ & (-30 \hat{\mathbf{i}}-24 \hat{\mathbf{j}}+12 \hat{\mathbf{k}}) \\ & =30-216+168=-18 \end{aligned} $

$ \begin{aligned}(d) & 3 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}+8 \hat{\mathbf{k}} ;(3 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}+8 \hat{\mathbf{k}}) \\ & (-30 \hat{\mathbf{i}}-24 \hat{\mathbf{j}}+12 \hat{\mathbf{k}}) \\ & =-90+168+96 \neq-18 \end{aligned} $

Hence, the correct option is (c).

$ \begin{aligned} &\text { Given, angle between the planes }\\ &\begin{aligned} & \mathbf{r} \cdot(11 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\alpha \hat{\mathbf{k}})=7 \text { and } \\ & \mathbf{r} \cdot(2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})=5 \text { is } \pi / 2 \\ & \therefore \cos \frac{\pi}{2}=\frac{11 \times 2+(-2) \times 4-2 \alpha}{\sqrt{121+4+\alpha^2} \sqrt{4+16+4}} \end{aligned} \end{aligned} $

$ \begin{aligned} \Rightarrow & & 0 & =22-8-2 \alpha \\ \Rightarrow & & 2 \alpha & =14 \\ \Rightarrow & & \alpha & =7 \end{aligned} $

Given, $A(27,-243,81)$ is point in space.

$B, C$ and $D$ are the images of $A$ with respect to $X Y, Y Z$ and $Z X$ plane, respectively.

∴ Coordinate of $B, C$ and $D$ are $(27,-243,-81),(-27,-243,81)$ and (27, 243, 81), respectively.

Centroid of $B C D$ is

$ \begin{aligned} & \left(\frac{27-27+27}{3}, \frac{-243-243+243}{3},\right. \left.\frac{-81+81+81}{3}\right) \\ & =(9,-81,27) \\ & \therefore \alpha=9, \beta=-81 \text { and } \gamma=27 \\ & \alpha+\beta+\gamma=9-81+27=-45 \end{aligned} $

Given, a point $A(2,5,7)$ be the image of the point $B(1,-2,3)$, w.r.t. plane $\pi$.

Mid-point of $A B$ is $C$.

Coordinate of $C$ is $\left(\frac{3}{2}, \frac{3}{2}, 5\right)$.

DR's of $C D$ are $2-\frac{3}{2}, 1-\frac{3}{2}, 6-5$

$ =\frac{1}{2},-\frac{1}{2}, 1 $

DC's of $C D$ are

$ \begin{aligned} & =\frac{\frac{1}{2}}{\sqrt{\frac{1}{4}+\frac{1}{4}+1}}, \frac{-\frac{1}{2}}{\sqrt{\frac{1}{4}+\frac{1}{4}+1}}, \frac{1}{\sqrt{\frac{1}{4}+\frac{1}{4}+1}} \\ & =\frac{1}{\sqrt{6}},-\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}} \end{aligned} $

Let the point $P$ divides the line $A B$ in the

ratio $\lambda: 1$.

Coordinate of $P$ is

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

Point $P$ lies on the plane

$ \begin{aligned} & x+y+z-5=0 \\ & \Rightarrow \frac{2 \lambda+1}{\lambda+1}+\frac{2 \lambda+1}{\lambda+1}+\frac{2 \lambda+1}{\lambda+1}=5 \\ & \Rightarrow \quad 6 \lambda+3=5 \lambda+5 \\ & \Rightarrow \quad \lambda=2 \end{aligned} $

∴ The point $P$ divides line $A B$ in the ratio 2:1.

Hence, $A P: P B=2: 1$

Let position vector of

$ \mathbf{A}=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}} $

Equation of line $L$ :

$ \frac{x-a}{2}=\frac{y-b}{1}=\frac{z-c}{-2} $

Point $P$ also lies on line $L$

$ \begin{aligned} & \frac{-7-a}{2}=\frac{-5-b}{1}=\frac{11-c}{-2}=\lambda(\text { let }) \\ & a=-7-2 \lambda, b=-5-\lambda \\and\,\, & c=11+2 \lambda \end{aligned} $

$ \begin{aligned} &\begin{aligned} \mathbf{A P} & =2 \lambda \hat{\mathbf{i}}+\lambda \hat{\mathbf{j}}-2 \lambda \hat{\mathbf{k}} \\ |\mathbf{A P}| & =\sqrt{4 \lambda^2+\lambda^2+4 \lambda^2}=12 \\ 3 \lambda & =12 \Rightarrow \lambda=4 \\ \therefore \quad a & =-15, b=-9 \text { and } c=19 \end{aligned}\\ &\text { Positive vector of } \mathbf{A} \text { is }-15 \hat{\mathbf{i}}-9 \hat{\mathbf{j}}+19 \hat{\mathbf{k}} \end{aligned} $

Equation of bisector of no mals of the plane is

$ \begin{aligned} & \frac{4 x+3 y+\lambda}{\sqrt{4^2+3^2}}= \pm \frac{x+2 y+2 z+d}{\sqrt{1+4+4}} \\ & \frac{4 x+3 y+\lambda}{5}= \pm \frac{x+2 y+2 z+d}{3} \\ & 12 x+9 y+3 \lambda= \pm(5 x+10 y+10 z+5 d) \end{aligned} $

The equations are

$ \begin{aligned} & 7 x-y-10 z+3 \lambda-5 d=0 \\ & \text { and } 17 x+19 y+10 z+3 \lambda+5 d=0 \end{aligned} $

Hence, the bisector of normal is either along $17 \hat{\mathbf{i}}+19 \hat{\mathbf{j}}+10 \hat{\mathbf{k}}$ and $7 \hat{\mathbf{i}}-\hat{\mathbf{j}}-10 \hat{\mathbf{k}}$.

Let coordinate of $D$ is $(a, b, c)$.

As $C$ is centroid of $A B C D$,

$ \begin{aligned} & G\left(\frac{1+2+3+a}{4},\right. \frac{2-3+2+b}{4} ,\left.\frac{3+1-1+c}{4}\right) \end{aligned} $

$ \begin{array}{ll} \Rightarrow & G\left(\frac{6+a}{4}, \frac{1+b}{4}, \frac{3+c}{4}\right) \\ \Rightarrow & \frac{6+a}{4}=\frac{5}{2}, \frac{1+b}{4}=\frac{3}{2}, \frac{3+c}{4}=\frac{9}{4} \\ \Rightarrow & a=4, b=5 \text { and } c=6 \end{array} $

∴ Coordinate of $G$ is $(4,5,6)$.

Let the point which divides $G D$ in the ratio $1: 2$ is ( $h, k, l$ ).

$ \begin{aligned} & \Rightarrow \quad h=\frac{1 \times 4+2 \times \frac{5}{2}}{1+2}, \\ & \quad k=\frac{1 \times 5+2 \times \frac{3}{2}}{1+2} \\ & \text { and } \quad l=\frac{1 \times 6+2 \times \frac{9}{4}}{1+2} \\ & \Rightarrow \quad h=3, k=\frac{8}{3} \text { and } l=\frac{7}{2} \end{aligned} $

∴ The required point is $\left(3, \frac{8}{3}, \frac{7}{2}\right)$.

$ \text { Let } D \text { divides } B C \text { in the ratio } k: 1 \text {. } $

Coordinate of $D$ is

$ \left(\frac{-2 k}{k+1}, \frac{4}{k+1}, \frac{4 k+1}{k+1}\right) $

Direction ratios of $B C=2,4,-3$

Direction ratios of $A D$ is $\frac{-2 k}{k+1}-2$,

$ \begin{aligned} & \frac{4}{k+1}, \frac{4 k+1}{k+1}-3 \\ & \quad=\frac{-4 k-2}{k+1}, \frac{4}{k+1}, \frac{k-2}{k+1} \end{aligned} $

As $A D \perp B C$,

$ \begin{gathered} 2\left(\frac{-4 k-2}{k+1}\right)+4\left(\frac{4}{k+1}\right)-3\left(\frac{k-2}{k+1}\right)=0 \\ -8 k-4 \pm 16-3 k+6=0 \\ -11 k+18=0 \Rightarrow k=18 / 11 \end{gathered} $

$\therefore D$ divides $B C$ in the ratio $18: 11$.

Given, equation of plane

$ \begin{aligned} & 6 x-3 y+2 z=6 \\ & \Rightarrow \quad x+\frac{y}{-2}+\frac{z}{3}=1 \\ & \therefore a=1, b=-2 \text { and } c=3 \end{aligned} $

Direction ratios of plane is $6,-3,2$.

Direction cosines of plane is

$ \begin{aligned} & \frac{6}{\sqrt{36+9+4}}, \frac{-3}{\sqrt{36+9+4}}, \frac{2}{\sqrt{36+9+4}} \\ & \therefore \quad l=\frac{6}{7}, m=\frac{-3}{7} \text { and } n=\frac{2}{7} \\ & \quad p=\left|\frac{0-0+0-6}{\sqrt{6^2+(-3)^2+2^2}}\right|=\frac{6}{7} \\ & \begin{aligned} |a l+b m+c n| & =\left|\frac{6}{7}+\frac{6}{7}+\frac{6}{7}\right|=3 \times \frac{6}{7} \\ & =3 p \end{aligned} \end{aligned} $

Equation of line $L$ is

$ \frac{x}{1}=\frac{y-3}{-1}=\frac{z-2}{1} $

Any point on this line can be

$ \begin{aligned} & (k,-k+3, k+2) \\ & \therefore \text { Let } A=(k,-k+3, k+2), B \equiv(0,3,2) \end{aligned} $

Now, $A B=3$

$ \Rightarrow \quad k^2+k^2+k^2=9 \Rightarrow k= \pm \sqrt{3} $

Normal to the plane is

$ \left|\begin{array}{ccc} \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}} $

DR's of normal are $\langle 1,1,-1\rangle$

Equation of normal is

$ \frac{x-\sqrt{3}}{1}=\frac{y+\sqrt{3}-3}{1}=\frac{z-\sqrt{3}-2}{-1} $

which is same as option (a).

$\pi_1$ is $\mathbf{r} \cdot(a \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}})=2 a-5$

$ \begin{aligned} & \pi_2 \text { is } \mathbf{r} \cdot(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}})=-4 \\ & \cos \theta=-\sqrt{\frac{3}{7}}=\frac{(a \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}})}{\sqrt{a^2+4+9} \sqrt{1+4+1}} \\ & \Rightarrow \quad-\sqrt{\frac{3}{7}} \times \sqrt{6}=\frac{(a-4-3)}{\sqrt{a^2+13}} \\ & \Rightarrow \quad \frac{(a-7)^2}{a^2+13}=\frac{18}{7} \\ & \Rightarrow \quad 7 a^2-98 a+343=18 a^2+234 \\ & \Rightarrow \quad 11 a^2+98 a-109=0 \\ & \Rightarrow \quad(11 a+109)(a-1)=0 \\ & \Rightarrow \quad a=1 \text { or } a=-109 / 11 \end{aligned} $

$\therefore \quad a=1$ is the integral solution.

$A(1,3,5), B(2,4,6), C(4,5, k)$

$ \begin{aligned} \Rightarrow \quad \mathbf{A B}=(1,1,1) ; \quad \mathbf{B C}=(2,1, k-6) \\ \quad \mathbf{A C}=(3,2, k-5) \end{aligned} $

According to the question, $\mathbf{A B} \cdot \mathbf{B C}=\mathbf{0}$

$ \begin{aligned} & \Rightarrow \quad 2+1+k-6=0 \Rightarrow k=3 \\ & \operatorname{AB} \cdot \mathbf{A C}=0 \\ & \Rightarrow \quad 3+2+k-5=0 \\ & \Rightarrow \quad k=0 \end{aligned} $

$\mathbf{B C} \cdot \mathbf{A C}=\mathbf{0}$

$ \begin{array}{ll} \Rightarrow & 6+2+(k-6)(k-5)=0 \\ \Rightarrow & k^2-11 k+38=0 \\ & D<0 \end{array} $

⇒ No real values of $k$ Hence, number of possible values of $k$ is 2 .

$ \text { } \begin{aligned} \mathbf{A B} & =\hat{\mathbf{i}}+4 \hat{\mathbf{k}} \\ \mathbf{C D} & =7 \hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}} \\ |\mathbf{A B}| & =\sqrt{1^2+4^2}=\sqrt{17} \\ |\mathbf{C D}| & =\sqrt{49+1+1}=\sqrt{51} \\ \therefore \quad \cos \theta & =\frac{\mathbf{A B} \cdot \mathbf{C D}}{|\mathbf{A B}||\mathbf{C D}|}=\frac{7-4}{\sqrt{17} \sqrt{51}}=\frac{3}{17 \sqrt{3}} \end{aligned} $

Vector perpendicular to the plane

$ \begin{aligned} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ -1 & 2 & 1 \\ 1 & 3 & 2 \end{array}\right|=\hat{\mathbf{i}}(1)-\hat{\mathbf{j}}(-2-1)+\hat{\mathbf{k}}(-3-2) \\ & =\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}} \end{aligned} $

∴ DR's of normal are $\langle 1,3,-5)\rangle$.

Equation of plane

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

∵ It passes through the point $(3,-1,4)$.

$ \begin{array}{lrl} \therefore & (3-2)+3(-1-1)-5(4-k) & =0 \\ \Rightarrow & 1-6-20+5 k & =0 \\ \Rightarrow & & k =5 \end{array} $

Equation of line $L: \mathbf{r}=\mathbf{a}+\lambda \mathbf{b}$

$ \mathbf{r}=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})+\lambda(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}) $

$P$ be any point on line $L$.

$P(1+2 \lambda, 2-3 \lambda,-2-6 \lambda)$ and $A(1,2,-2)$

Given, $A P=21$

$ \begin{aligned} & \sqrt{(2 \lambda)^2+(-3 \lambda)^2+(-6 \lambda)^2}=21 \\ & \Rightarrow|\lambda|(\sqrt{4+9+36})=21 \\ & \Rightarrow|\lambda|=3 \Rightarrow \lambda= \pm 3 \end{aligned} $

At $\lambda=3, P(7,-7,-20)$

Position vector of $P: 7 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-20 \hat{\mathbf{k}}$

At $\lambda=-3$ and $P(-5,11,16)$

Position vector of $P:-5 \hat{\mathbf{i}}+11 \hat{\mathbf{j}}+16 \hat{\mathbf{k}}$

Given, point on plane $(1,-2,3)$

$ a=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} $

Perpendicular vector perpendicular to plane

$ -\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} $

Equation of required plane, $\mathbf{r} \cdot \mathbf{n}=\mathbf{a} \cdot \mathbf{n}$

$ \begin{aligned} & (x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \cdot(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}) \\ = & (\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \cdot(-\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}) \\ \Rightarrow \quad & x+2 y-3 z=-1-4-9=-14 \\ \Rightarrow \quad & x-2 y+3 z=14 \end{aligned} $

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} & A(1,2,3), B(-1,4,6), C(0,-6,4), D(1,1,1) \\ & G\left(\frac{1-1+0+1}{4}, \frac{2+4-6+1}{4}\right. , \left.\frac{3+6+4+1}{4}\right) \equiv G\left(\frac{1}{4}, \frac{1}{4}, \frac{14}{4}\right) \\ & G_1\left(\frac{-1+0+1}{3}, \frac{4-6+1}{3}, \frac{6+4+1}{3}\right) \\ & \equiv G_1\left(0, \frac{-1}{3}, \frac{11}{3}\right) \\ & A G_1=\sqrt{1^2+\left(\frac{7}{3}\right)^2+\left(\frac{2}{3}\right)^2} \\ & \quad=\frac{1}{3} \sqrt{9+49+4}=\frac{\sqrt{62}}{3} \\ & A G=\sqrt{\left(\frac{3}{4}\right)^2+\left(\frac{7}{4}\right)^2+\left(\frac{2}{4}\right)^2}=\frac{\sqrt{62}}{4} \\ & \frac{A G_1}{A G}=\frac{\sqrt{62} / 3}{\sqrt{62} / 4}=\frac{4}{3} \end{aligned} \end{aligned} $

Here, $P_1: x-y+z+2=0$

$ \begin{aligned} & n_1: 1,-1,1 \\ & P_2: 2 x+y-2 z+5=0 \\ & n_2: 2,1,-2 \end{aligned} $

vector along line of intersection of $P_1$ and $P_2=\mathbf{n}_1 \times \mathbf{n}_2$

$ \begin{aligned} & \mathrm{n}=\left|\begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 1 \\ 2 & 1 & -2 \end{array}\right|=\mathbf{i}+4 \mathbf{j}+3 \mathbf{k} \\ & |\mathrm{n}|=\sqrt{1+16+9}=\sqrt{26} \\ & \text { DC's are } \frac{1}{\sqrt{26}}, \frac{4}{\sqrt{26}}, \frac{3}{\sqrt{26}} \end{aligned} $

$ \text { Dr's of normal to plane } $

$ \begin{aligned} & : 1-(-1), 2-3,3-(-2) \\ & :(2,-1,5) \equiv(a, b, c) \end{aligned} $

Plane passes through

$ (-1,3,-2) \equiv\left(x_1, y_1, z_1\right) $

Equation of plane

$ \begin{array}{r} a\left(x-x_1\right)+b\left(y-y_1\right)+c\left(z-z_1\right)=0 \\ 2(x+1)-(y-3)+5(z+2)=0 \\ 2 x-y+5 z+15=0 \end{array} $

Perpendicular distance from origin to the plane

$ \left|\frac{0-0+0+15}{\sqrt{4+1+25}}\right|=\frac{15}{\sqrt{30}}=\sqrt{\frac{15}{2}} $

Given that, points $A(4,7,8), B(2,3,4)$ and $C(2,5,7)$

Length

$ \begin{aligned} A B & =\sqrt{\left(x_A-x_B\right)^2+\left(y_A-y_B\right)^2+\left(z_A-z_B\right)^2} \\ & =\sqrt{(4-2)^2+(7-3)^2+(8-4)^2} \end{aligned} $

$ \begin{aligned} & =\sqrt{(2)^2+(4)^2+(4)^2}=\sqrt{4+16+16} \\ A B & =6 \end{aligned} $

Length

$ \begin{aligned} A C & =\sqrt{\left(x_A-x_C\right)^2+\left(y_A-y_C\right)^2+\left(z_A-z_C\right)^2} \\ & =\sqrt{(4-2)^2+(7-5)^2+(8-7)^2} \\ & =\sqrt{(2)^2+(2)^2+(1)^2}=\sqrt{4+4+1} \end{aligned} $

$ A C=3 \Rightarrow A B: A C=6: 3=2: 1 $

Now, let say the internal bisector of the angle $A$ meets the side $B C$ that point

$ D\left(x_D, y_D, z_D\right) $

$\therefore x$-coordinate of point $D$,

$ x_D=\frac{m x_C+n x_B}{m+n}=\frac{2 \times 2+1 \times 2}{2+1} $

$ \Rightarrow x_D=2 $

$y$-coordinate of point $D$,

$ y_D=\frac{m y_C+n y_B}{m+n}=\frac{2 \times 5+1 \times 3}{2+1}=\frac{13}{3} $

$z$-coordinate of point $D$,

$ \begin{aligned} & z_D=\frac{m z_C+n z_B}{m+n}=\frac{2 \times 7+1 \times 4}{2+1}=\frac{18}{3}=6 \\ & \because \text { Point } D\left(2, \frac{13}{3}, 6\right) \end{aligned} $

So, length

$ \begin{aligned} & =\sqrt{(4-2)^2+\left(7-\frac{13}{3}\right)^2+(8-6)^2} \\ & =\sqrt{(2)^2+\left(\frac{8}{3}\right)^2+(2)^2} \\ & =\sqrt{4+\frac{64}{9}+4}=\sqrt{\frac{136}{9}}=\frac{2}{3} \sqrt{34} \end{aligned} $

Given, vector equation of plane

$ \begin{align*} & \mathbf{r}=(2+3 \lambda-\mu) \hat{\mathbf{i}}+(1-2 \lambda+3 \mu) \\ & \hat{\mathbf{j}}+(-2+2 \lambda+\mu) k \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{align*} $

Let $\mathbf{r}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

∴ Compare Eqs. (i) and (ii),

$ \begin{gather*} 2+3 \lambda-\mu=x \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\\ 1-2 \lambda+3 \mu=y \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\\ -2+2 \lambda+\mu=z \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v) \end{gather*} $

Now, adding Eqs. (iii) and (iv),

$ \begin{align*} & 5 \lambda=x+z \\ & \Rightarrow \lambda=\frac{1}{5}(n+z) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(vi) \end{align*} $

On adding Eqs. (iv) and (vi)

$ \begin{align*} & & -1+4 \mu & =y+z \\ \Rightarrow & & \mu & =\frac{1}{4}(y+z+1) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(vii) \end{align*} $

$ \begin{aligned} &\text { Putting value of } \lambda \text { and } \mu \text { in Eq. (iii), }\\ &\begin{aligned} & & 2+\frac{3}{5}(x+z)-\frac{1}{4}(y+z+1) & =x \\ & \Rightarrow & 40+12(x+z)-5(y+z+1) & =20 x \\ & \Rightarrow & 40+12 x+12 z-5 y-5 z-5 & =20 x \\ & \Rightarrow & -8 x-5 y+7 z+35 & =0 \\ & \Rightarrow & 8 x+5 y-7 z-35 & =0 \end{aligned} \end{aligned} $

Given projections of $A$ and $B$ are $P$ and $Q$ respectively

$ A(1,2,0) \text { and } B(2,1,1) $

Plane $P: x+y+z=3$

Vector perpendicular to plane $P$ : is

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

∴ vector equation of line $A P$ and $B Q$ are

$ \begin{gathered} A P: \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\lambda(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}), B Q: 2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\mathbf{k} \\ +\mu(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \end{gathered} $

∴ Any point on line $A P$ and $B Q$ is

$ P(1+\lambda, 2+\lambda, \lambda), Q(2+\mu, 1+\mu, 1+\mu) $

Point $P$ and $Q$ lies on the plane $x+y+z=3$

$ \therefore \quad 1+\lambda+2+\lambda+\lambda=3 \Rightarrow \lambda=0 $

and $2+\mu+1+\mu+1+\mu=3$

$ \Rightarrow \quad \mu=\frac{-1}{3} $

Point $P(1,2,0)$,

$ \begin{aligned} & Q\left(2-1-\left(\frac{-1}{3}\right), 1-\frac{1}{3}, 1-\frac{1}{3}\right) \\ & Q\left(\frac{5}{3}, \frac{2}{3}, \frac{2}{3}\right) \end{aligned} $

$ \begin{aligned} &\text { ∴ Distance between }\\ &\begin{aligned} P Q & =\sqrt{\left(\frac{5}{3}-1\right)^2+\left(\frac{2}{3}-2\right)^2+\left(\frac{2}{3}-0\right)^2} \\ P Q & =\sqrt{\frac{4}{9}+\frac{16}{9}+\frac{4}{9}} \Rightarrow P Q=\sqrt{\frac{24}{9}} \\ \Rightarrow P Q & =\frac{2 \sqrt{2}}{\sqrt{3}} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Given points are }\\ &\begin{aligned} & A(4,3,2), B(5,4,6), C(-1,-1,5) \\ & \begin{aligned} A B & =\sqrt{(5-4)^2+(4-3)^2+(6-2)^2} \\ & =\sqrt{(1)^2+(1)^2+(4)^2}=\sqrt{18} \\ A B & =3 \sqrt{2} \end{aligned} \end{aligned} \end{aligned} $

$ \begin{aligned} A C & =\sqrt{(-1-4)^2+(-1-3)^2+(5-2)^2} \\ & =\sqrt{(-5)^2+(-4)^2+(3)^2} \\ & =\sqrt{25+16+9}=\sqrt{50} \\ A C & =5 \sqrt{2} \\ A B & : A C=3 \sqrt{2}: 5 \sqrt{2}=3: 5 \end{aligned} $

Let say the bisector of angle $A$ meet the side $B C$ that point $D\left(x_D, y_D, z_D\right)$ and $D$ divides $B C$ side in ratio 3: 5 internally

$ \begin{aligned} & x_D=\frac{m x_C+n x_B}{m+n}=\frac{3 \times(-1)+5 \times 5}{3+5}=\frac{22}{8} \\ & y_D=\frac{m y_C+n y_B}{m+n}=\frac{3 \times(-1)+5 \times(4)}{3+5}=\frac{17}{8} \\ & z_D=\frac{m z_C+n z_B}{m+n}=\frac{3 \times(5)+5 \times(6)}{3+5}=\frac{45}{8} \end{aligned} $

Therefore, coordinates of point $D$ is $\left(\frac{22}{8}, \frac{17}{8}, \frac{45}{8}\right)$.

Let the direction ratios of a line $L_1$ be $a_1, b_1, c_1$ and those of another line $L_2$ are $a_2, b_2, c_2$ and the angle $\theta$ between them is given by

$ \begin{equation*} \cos \theta=\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{equation*} $

If both lines are parallel then, their direction ratios must be in equal proportion.

$i.e., \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} $

$ \begin{aligned} &\text { Now, from Eq. (i) }\\ &\begin{aligned} & \cos \theta=\frac{(2)\left(\frac{4}{\sqrt{19}}\right)+(5)\left(\frac{10}{\sqrt{19}}\right)+(7)\left(\frac{14}{\sqrt{19}}\right)}{\sqrt{2^2+5^2+7^2}} \\ & \sqrt{\begin{array}{c} \left(\frac{4}{\sqrt{19}}\right)^2+\left(\frac{10}{\sqrt{19}}\right)^2 \\ +\left(\frac{14}{\sqrt{19}}\right)^2 \end{array}} \\ & =\frac{\frac{8}{\sqrt{19}}+\frac{50}{\sqrt{19}}+\frac{98}{\sqrt{19}}}{\sqrt{78} \sqrt{\frac{312}{19}}}=\frac{\frac{106}{\sqrt{19}}}{\frac{78 \times 2}{\sqrt{19}}}=1 \\ & \Rightarrow \theta=0 \\ & \text { or } \frac{a_1}{a_2}=\frac{2}{\frac{4}{\sqrt{19}}}=\frac{\sqrt{19}}{2} \end{aligned} \end{aligned} $

$ \begin{array}{ll} \Rightarrow & \frac{b_1}{b_2}=\frac{5}{\frac{10}{\sqrt{19}}}=\frac{\sqrt{19}}{2} \\ & \frac{c_1}{c_2}=\frac{7}{\frac{14}{\sqrt{19}}}=\frac{\sqrt{19}}{2} \\ \because & \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} \end{array} $

Hence, both lines are parallel to each other.

Given that, line

$ \frac{x-4}{1}=\frac{y-2}{1}=\frac{z-7}{2} $

Line is passing through point $P(4,+2,7)$ and direction ratios ( $1,1,2$ ) and lies on the given plane $a x+b y+z=7$

So, point will satisfy the equation of plane $a(4)+b(+2)+(7)=7$

$ \begin{equation*} 4 a+2 b=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{equation*} $

Now direction ratios of the line and direction ratios of the plane will be perpendicular so their product must be zero

$ \begin{aligned} a_1 a_2+b_1 b_2 & +c_1 c_2 \\ & =(1)(a)+(1)(b)+(2)(1) \\ 0 & =a+b+2 \end{aligned} $

or    $ a+b=-2 $

Given, lines

$ \begin{aligned} \mathbf{r} & =(-\hat{\mathbf{i}}+3 \hat{\mathbf{k}})+t(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}) \\ \text { and } \mathbf{r} & =(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})+s(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \end{aligned} $

Shortest distance between lines is

$ \begin{aligned} & ((3+1) \hat{\mathbf{i}}+\hat{\mathbf{j}}-(\hat{\mathbf{k}}+3) \hat{\mathbf{k}}) \\ S D & =\frac{-[(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}) \times(2 \hat{\mathbf{i}}-\mathbf{j}+2 \mathbf{k})]}{|(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})| \times(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \mid} \\ S D & =\frac{(4 \hat{\mathbf{i}}+\hat{\mathbf{j}}-4 \hat{\mathbf{k}}) \cdot(12 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}-8 \hat{\mathbf{k}})}{|12 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}-8 \hat{\mathbf{k}}|} \\ S D & =\frac{48+8+32}{\sqrt{144+64+64}}=\frac{88}{4 \sqrt{17}}=\frac{22}{\sqrt{17}} \end{aligned} $

Direction ratio of diagonal $d_1$ i.e., direction of $A E$ is $a \hat{\mathbf{i}}-b \hat{\mathbf{j}}$

Direction ratio of diagonal $d_2$ i.e., direction of $A D a \hat{\mathbf{i}}-c \hat{\mathbf{k}}$

Angle between diagonal $d_1$ and diagonal $d_2$

$ \begin{aligned} \cos \theta & =\frac{(a \hat{\mathbf{i}}-b \hat{\mathbf{j}}) \cdot(a \hat{\mathbf{i}}-c \hat{\mathbf{k}})}{\sqrt{a^2+b^2} \sqrt{a^2+c^2}} \\ & =\frac{a^2}{\sqrt{a^2+b^2} \sqrt{a^2+c^2}} \\ \Rightarrow \quad \theta & =\cos ^{-1}\left(\frac{a^2}{\sqrt{a^2+b^2} \sqrt{a^2+c^2}}\right) \end{aligned} $

$ \begin{aligned} & \text {We have, } a+b+c=0 \\ & \text { and } 2 a b+2 a c-b c=0 \\ & \quad a=-(b+c) \\ & \therefore \quad-2(b+c) b+-2(b+c) c-b c=0 \\ & \Rightarrow \quad-2 b^2-2 b c-2 b c-2 c^2-b c=0 \\ & \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\quad 2 b^2+5 b c+2 c^2=0 \\ & \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\quad 2 b^2+4 b c+b c+2 c^2=0 \\ & \Rightarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \quad(2 b+c)(b+2 c)=0 \\ & \Rightarrow \quad b=-2 c, b=-c / 2 \\ & \quad a=-(-2 c+c)=c \\ & \Rightarrow \quad a=-\left(\frac{-c}{2}+c\right)=\frac{-c}{2} \\ & \therefore a_1=1, b_1=-2, c_1=1, \\ & \quad a_2=\frac{-1}{2}, b_2=\frac{-1}{2}, c_2=1 \\ & \quad a_1=1, b_1=-2, c_1=1, \\ & a_2=1, b_2=1, c_2=-2 \\ & \cos \theta=\frac{(1)(1)+(-2)(1)+(1)(-2)}{\sqrt{1+4+1} \sqrt{1+1+4}} \\ & \cos \theta=\frac{-1}{2} \Rightarrow \theta=\frac{2 \pi}{3} \end{aligned} $

Let the equation of plane is, $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$

where $a, b, c$ are $x, y$ and $z$ intercepts ∴ Centroid of $\triangle A B C$ is $\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)$.

Given, $\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)=(2,3,5)$

$ \therefore \quad a=6, b=9, c=15 $

Equation of plane is, $\frac{x}{6}+\frac{y}{9}+\frac{z}{15}=1$

$ 15 x+10 y+6 z=90 $

Equation of plane ( $\pi$ ) containing the points

$ \begin{aligned} & (0,-5,-1),(1,-2,5),(-3,5,0) \\ & \text { is }\left|\begin{array}{ccc} x-0 & y+5 & z+1 \\ 1-0 & -2+5 & 5+1 \\ -3-0 & 5+5 & 0+1 \end{array}\right|=0 \\ & \Rightarrow \quad\left|\begin{array}{ccc} x & y+5 & z+1 \\ 1 & 3 & 6 \\ -3 & 10 & 1 \end{array}\right|=0 \\ & \Rightarrow \quad 3 x+y-z+4=0 \\ & \Rightarrow \quad \mathbf{r}(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})+4=0 \\ & \Rightarrow \quad \mathbf{r}(-3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})=4 \\ & \Rightarrow \mathbf{r}\left(\frac{-3}{\sqrt{11}}-\frac{\hat{\mathbf{j}}}{\sqrt{11}}+\frac{\hat{\mathbf{k}}}{\sqrt{11}}\right)=\frac{4}{\sqrt{11}} \end{aligned} $

∴ Unit normal vector to plane $\pi$ is

$ \hat{\mathbf{n}}=\frac{-3}{\sqrt{11}} \hat{\mathbf{i}}-\frac{\hat{\mathbf{j}}}{\sqrt{11}}+\frac{\hat{\mathbf{k}}}{\sqrt{11}} $

∴ length of the projection of the unit normal vector $\hat{\mathbf{n}}$ to plane $\pi$ on the line $L$ is

$ \left|\frac{\left(\frac{-3}{\sqrt{11}} \hat{\mathbf{i}} \frac{-\hat{\mathbf{j}}}{\sqrt{11}} \frac{+\hat{\mathbf{k}}}{\sqrt{11}}\right) \cdot(\hat{\mathbf{i}}+5 \hat{\mathbf{j}}-\hat{\mathbf{k}})}{\sqrt{1+25+36}}\right| $

$ =\left|\frac{\frac{-3}{\sqrt{11}} \frac{-5}{\sqrt{11}} \frac{-6}{\sqrt{11}}}{\sqrt{62}}\right|=\left|\frac{-14}{\sqrt{11} \sqrt{62}}\right|=\frac{14}{\sqrt{682}} $

Equation of line passing through the points $(a, 2,-4)$ and $(5,3, b)$ is given by

$ \begin{aligned} \frac{x-a}{5-a} & =\frac{y-2}{3-2}=\frac{z+4}{b+4} \\ \Rightarrow \quad \frac{x-a}{5-a} & =\frac{y-2}{1}=\frac{z+4}{b+4} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

The equation of $Z X$-plane, $y=0$ Since, line (i) meets the $Z X$-plane.

$ \begin{array}{ll} \Rightarrow & \frac{x-a}{5-a}=\frac{0-2}{1}=\frac{z+4}{b+4} \\ \Rightarrow & \frac{x-a}{5-a}=-2 \text { and } \frac{z+4}{b+4}=-2 \\ \Rightarrow & x-a=-2(5-a) \\ \text { and } & z+4=-2(b+4) \end{array} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad x=a-10+2 a \\ & \text { and } z+4=-2 b-8 \\ & \Rightarrow \quad x=3 a-10 \text { and } z=-2 b-12 \\ & \text { and } y=0 \\ & \therefore \quad 3 a-10=-a+2 b \text { and } \\ & -2 b-12=a+b \\ & \Rightarrow \quad 4 a-2 b=10 \text { and } a+3 b=-12 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & \Rightarrow \quad 2 a-b=5 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned}\\ &\text { On solving Eqs (i) and (ii), we get }\\ &\begin{aligned} a=\frac{3}{7} & \text { and } b=\frac{-29}{7} \\ \therefore \quad 14 a+7 b & =14\left(\frac{3}{7}\right)+7\left(\frac{-29}{7}\right) \\ & =6-29=-23 \end{aligned} \end{aligned} $