(i) (b): The line along which motorcycle A is running, \(\vec{r}=\lambda(\hat{i}+2 \hat{j}-\hat{k})\) is which can be rewritten as \((x \hat{i}+y \hat{j}+z \hat{k})=\lambda \hat{i}+2 \lambda \hat{j}-\lambda \hat{k}\)
\(\Rightarrow x=\lambda, y=2 \lambda, z=-\lambda \Rightarrow \frac{x}{1}=\lambda, \frac{y}{2}=\lambda, \frac{z}{-1}=\lambda\)
Thus, the required cartesian equation is \(\frac{x}{1}=\frac{y}{2}=\frac{z}{-1}\)
(ii) (d): Clearly, D.R:s of the required line are <  1, 2, -1  >
∴ D.Cs are \( <\frac{1}{\sqrt{1^{2}+2^{2}+(-1)^{2}}}, \frac{2}{\sqrt{1^{2}+2^{2}+(-1)^{2}}}, \frac{-1}{\sqrt{1^{2}+2^{2}+(-1)^{2}}}> \)
\(\text { i.e., }<\frac{1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{-1}{\sqrt{6}}>\)
(iii) (d): The line along which motorcycle B is running, is \(\vec{r}=(3 \hat{i}+3 \hat{j})+\mu(2 \hat{i}+\hat{j}+\hat{k})\), which is parallel to the vector \(2 \hat{i}+\hat{j}+\hat{k}\).
∴ D.R.'s of the required line are < 2, 1, 1 >.
(iv) (d): Here, \(\vec{a}_{1}=0 \hat{i}+0 \hat{j}+0 \hat{k}, \vec{a}_{2}=3 \hat{i}+3 \hat{j}, \vec{b}_{1}=\hat{i}+2 \hat{j}-\hat{k} \vec{b}_{2}=2 \hat{i}+\hat{j}+\hat{k}\)
\(\therefore \vec{a}_{2}-\vec{a}_{1}=3 \hat{i}+3 \hat{j}\)
and  \(\vec{b}_{1} \times \vec{b}_{2}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 2 & 1 & 1 \end{array}\right|=3 \hat{i}-3 \hat{j}-3 \hat{k}\) 
Now, \(\left(\vec{a}_{2}-\vec{a}_{1}\right) \cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right)=(3 \hat{i}+3 \hat{j}) \cdot(3 \hat{i}-3 \hat{j}-3 \hat{k})\)
= 9 - 9 = 0.
Hence, shortest distance between the given lines is 0.
(v) (c): Since, the point (1, 2, -1) satisfy both the equations of lines, therefore point of intersection of given lines is (1, 2, -1). So, the motorcycles will meet with an accident at the point (1, 2, -1).

(i) (c): Clearly, the plane for students of school A is \(\vec{r} \cdot(\hat{i}+\hat{j}+2 \hat{k})=5,\) which can be rewritten as
\((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+\hat{j}+2 \hat{k})=5\)
\(\Rightarrow \quad x+y+2 z=5\), which is the required cartesian equation.
(ii) (b): Clearly, the equation of plane for students of school B is \(\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})=6,\) which is of the form \(\vec{r} \cdot \vec{n}=d\)
∴ Normal vector to the plane is, \(\vec{n}=2 \hat{i}-\hat{j}+\hat{k}\) and its magnitude is \(|\vec{n}|=\sqrt{2^{2}+(-1)^{2}+1^{2}}=\sqrt{6}\)
(iii) (b): The cartesian form is 2x - y + z = 6, which can be rewritten as
\(\frac{2 x}{6}-\frac{y}{6}+\frac{z}{6}=1 \Rightarrow \frac{x}{3}+\frac{y}{(-6)}+\frac{z}{6}=1\)
(iv) (c): Since, only the point (3, I, 1) satisfy the equation of plane representing seating position of students of school B, therefore Khushi is the student of school B.
(v) (d): Equation of plane representing students of school B is \(\vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})=6,\) which is not in normal form, as \(|\vec{n}| \neq 1\)
On dividing both sides by \(\sqrt{2^{2}+(-1)^{2}+1^{2}}=\sqrt{6}\), we get \(\vec{r} \cdot\left(\frac{2}{\sqrt{6}} \hat{i}-\frac{1}{\sqrt{6}} \hat{j}+\frac{1}{\sqrt{6}} \hat{k}\right)=\frac{6}{\sqrt{6}}\)
which is of the form \(\vec{r} \cdot \hat{n}=d\)
Thus, the required distance is \(\sqrt 6\) units.

(i) (b): Clearly, the coordinates of A are (8, -6, 0) and that of E are (0, 0, 24).
Also, cartesian equation of line along EA is given by
\(\frac{x-0}{8-0}=\frac{y-0}{-6-0}=\frac{z-24}{0-24}\)
\(\Rightarrow \frac{x}{8}=\frac{y}{-6}=\frac{z-24}{-24} \Rightarrow \frac{x}{-4}=\frac{y}{3}=\frac{z-24}{12}\)
(ii) (c): Clearly, the coordinates of Dare (-8, -6, 0) and that of E are (0, 0, 24)
\(\therefore \text { Vector } \overline{E D} \text { is }(-8-0) \hat{i}+(-6-0) \hat{j}+(0-24) \hat{k} \ \text { i.e., }-8 \hat{i}-6 \hat{i}-24 \hat{k} \)
(iii) (b): Since, the coordinates of Bare (8, 6, 0) and that of E are (0, 0, 24), therefore length of cable
\(E B=\sqrt{(8-0)^{2}+(6-0)^{2}+(0-24)^{2}} \)
\(=\sqrt{64+36+576}=\sqrt{676}=26 \text { units }\)
(iv) (d): Since, the coordinates of Care (-8,6,0) therefore length of cable EC = \(\sqrt{(-8-0)^{2}+(6-0)^{2}+(0-24)^{2}}\)
\(=\sqrt{64+36+576}=\sqrt{676}=26 \mathrm{units}\)
Similarly, length of cable EA = ED = 26 units.
(v) (c): Sum of all vectors along the cables
\(=\overrightarrow{E A}+\overrightarrow{E B}+\overrightarrow{E C}+\overline{E D}\)
\( =(8 \hat{i}-6 \hat{j}-24 \hat{k})+(8 \hat{i}+6 \hat{j}-24 \hat{k})+(-8 \hat{i}+6 \hat{j}-24 \hat{k})+(-8 \hat{i}-6 \hat{j}-24 \hat{k}) \)
\(=-96 \hat{k}\)

(i) (d) : Clearly, the coordinates of A are (8,10, 0) and Dare (0, 0, 30)
∴ Equation of AD is given by
\(\frac{x-0}{8-0}=\frac{y-0}{10-0}=\frac{z-30}{-30} \)
\(\Rightarrow \quad \frac{x}{4}=\frac{y}{5}=\frac{30-z}{15}\)
(ii) (b): The coordinates of point Care (15, -20, 0) and Dare (0, 0, 30)
∴ Length of the cable DC
\(=\sqrt{(0-15)^{2}+(0+20)^{2}+(30-0)^{2}}\)
\(=\sqrt{225+400+900}=\sqrt{1525}=5 \sqrt{61} \mathrm{~m}\)
(iii) (a) : Since, the coordinates of point Bare (-6, 4, 0) and Dare (0, 0, 30), therefore vector DB is
\((-6-0) \hat{i}+(4-0) \hat{j}+(0-30) \hat{k}, \text { i.e., }-6 \hat{i}+4 \hat{j}-30 \hat{k}\)
(iv) (b) : Required sum
\(=(\hat{i}+10 \hat{j}-30 \hat{k})+(-6 \hat{i}+4 \hat{j}-30 \hat{k})+(15 \hat{i}-20 \hat{j}-30 \hat{k}) \)
\(=17 \hat{i}-6 \hat{i}-90 \hat{k}\)
(v) (a): Clearly, OA = \(\sqrt{8^{2}+10^{2}}=\sqrt{164}\)
\(O B=\sqrt{6^{2}+4^{2}}=\sqrt{36+16}=\sqrt{52} \ \text { and } O C=\sqrt{15^{2}+20^{2}}=\sqrt{225+400}=\sqrt{625} \)

(i) (b) : Equation of plane is x - 2y + 2z = 3
∴ D.R's of normal to the plane are < 1, - 2, 2 >, which is also the D.R.'s of perpendicular from the point (3, -2, 1) to the given plane.
(ii) (c) : Required length = Perpendicular distance from (3, -2, 1) to the plane x - 2y + 2z = 3
\(=\left|\frac{3-2(-2)+2(1)-3}{\sqrt{1^{2}+(-2)^{2}+2^{2}}}\right|=\frac{6}{3}=2 \text { units }\)
(iii) (b) : The equation of perpendicular from the point (x₁, y₁, z₁) to the plane ax + by + cz = d is given by
\(\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\)
Here, x₁ = 3, y₁ = -2,  z₁= 1 and a = 1, b = -2, c = 2
∴ Required equation is \(\frac{x-3}{1}=\frac{y+2}{-2}=\frac{z-1}{2}\)
(iv) (d) : The equation of the plane parallel to the plane x - 2y + 2z - 3 = 0 is x - 2y + 2z + \(\lambda\) = 0
Now, distance of this plane from the point (3, -2, 1) is
\(\left|\frac{3+4+2+\lambda}{\sqrt{1^{2}+(-2)^{2}+2^{2}}}\right|=\left|\frac{9+\lambda}{3}\right|\)
But, this distance is given to be unity
\(\therefore|9+\lambda|=3 \Rightarrow \lambda+9=\pm 3 \Rightarrow \lambda=-6 \text { or }-12\)
Thus, required equation of planes are
x - 2y + 2z - 6 = 0 or x- 2y + 2z - 12 = 0
(v) (a): Let the coordinate of image of (3, -2, 1) be Q(r + 3, -2r - 2, 2r + 1)
Let R be the mid-point of PQ, then coordinate of R be \(\left(\frac{r+6}{2}, \frac{-2 r-4}{2}, r+1\right)\)
Since, R lies on the plane x - 2y + 2z = 3
\( \therefore \quad\left(\frac{r+6}{2}\right)-2\left(\frac{-2 r-4}{2}\right)+2(r+1)=3 \)
\(\Rightarrow \quad 9 r=-12 \Rightarrow r=-\frac{4}{3}\)
Thus, the coordinates of Q be \(\left(\frac{5}{3}, \frac{2}{3}, \frac{-5}{3}\right)\)

(i) (c) : Let P(2, 3, 5) and Q(1, 4, 2) be the positions of helicopter and boat respectively.
Now, direction ratios of PQ are proportional to 1-2, 4-3, 2-5, i.e , -1, 1,-3.
So, equation of plane passing through Q(1, 4, 2) and perpendicular to PQ is
-(x - 1) + (y - 4) + (-3) (z - 2) = 0 ⇒ x - y + 3z = 3
(ii) (d) : Required distance = Distance between P and Q
\(=\sqrt{(1-2)^{2}+(4-3)^{2}+(2-5)^{2}}=\sqrt{1+1+9}=\sqrt{11} \mathrm{~m}\)
(iii) (d) : We know, Distance = Speed x Time
ஃ Required time = \(\frac{\sqrt{11}}{30}\) seconds
(iv) (b) : Equation of line PQ is \(\frac{x-1}{1}=\frac{y-4}{-1}=\frac{z-2}{3}\)
(v) (b) : Any point on the line \(\frac{x-1}{1}=\frac{y-1}{-2}=\frac{z-2}{3}\) given by \((\lambda+1,-2 \lambda+1,3 \lambda+2)\)
Now, on substituting this point in the equation of plane x - y + 3z = 3, we get
\( (\lambda+1)-(-2 \lambda+1)+3(3 \lambda+2)=3 \)
\(\Rightarrow \lambda+1+2 \lambda-1+9 \lambda+6=3 \Rightarrow 12 \lambda=-3\)
\(\Rightarrow \ \lambda=\frac{-1}{4}\)
Thus, the required point is \(\left(\frac{-1}{4}+1, \frac{1}{2}+1, \frac{-3}{4}+2\right)\) i.e.,\(\left(\frac{3}{4}, \frac{3}{2}, \frac{5}{4}\right)\)
 

(i) (b) : D.R:s of OA are < 1-0, 0-0, 0-0 >, i.e., < 1, 0, 0 >.
(ii) (a) : Equation of diagonal OB' is \(\frac{x-0}{1}=\frac{y-0}{2}=\frac{z-0}{3} \text { i.e., } \frac{x}{1}=\frac{y}{2}=\frac{z}{3}\)
(iii) (c) : OABC is xy-plane, therefore its equation is z = 0.
(iv) (c) : Plane O'A'B'C is parallel to xy-plane passing through (0, 0, 3), therefore its equation is z = 3.
(v) (a) : Plane ABB' A' is parallel to yz-plane passing through (1, 0, 0), therefore its equation is x = 1.

(i) (a): Eliminating 't' from the given equations, we get equation of path as, \(\frac{x}{2}=\frac{y}{-4}=\frac{z}{4} \text { or } \frac{x}{1}=\frac{y}{-2}=\frac{z}{2}\).
Thus, the path of the rocket represents a straight line.
(ii).(b) : Since, only (1, -2, 2) satisfy the equation of path of rocket therefore (1, -2, 2) lie on the path of rocket.
(iii) (b): For t = 10 sec, we have x = 20,y = -40, z = 40 
Now, required distance = \(\sqrt{x^{2}+y^{2}+z^{2}}\)
\(=\sqrt{20^{2}+(-40)^{2}+(40)^{2}}=\sqrt{400+1600+1600}\)
\(=\sqrt{3600}=60 \mathrm{~km}\)
(iv) (d) : Clearly, height of rocket from the ground = z-coordinate of given position = 6 km
(v) (b) : Let Q be the image of point P(1, -2, 2) in the plane 3x - y + 4z = 2. Then, equation of PQ is
\(\frac{x-1}{3}=\frac{y+2}{-1}=\frac{z-2}{4}\)
Let the coordinates of Q be (3r +1, -r - 2, 4r + 2).
Let R be the mid -point of PQ. Then, coordinates of R are
\(\left(\frac{3 r+2}{2}, \frac{-r-4}{2}, \frac{4 r+4}{2}\right) \text { or }\left(\frac{3}{2} r+1, \frac{-r}{2}-2,2 r+2\right)\)
Since, R lies on 3x - y + 4z = 2.
\( \therefore \quad 3\left(\frac{3}{2} r+1\right)-\left(\frac{-r}{2}-2\right)+4(2 r+2)=2 \)
\(\Rightarrow \frac{9 r}{2}+3+\frac{r}{2}+2+8 r+8=2 \\ \)
\(\Rightarrow 13 r+13=2 \Rightarrow r=\frac{-11}{13}\)
Hence, the coordinates of Q are
\(\left(\frac{-33}{13}+1, \frac{11}{13}-2, \frac{-44}{13}+2\right) \text { i.e., }\left(\frac{-20}{13}, \frac{-15}{13}, \frac{-18}{13}\right)\)

(i) (a) : Since, D.R.'s are proportional to D.C.'s, therefore L; will be perpendicular to L₂ iff
\(l_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}=0\)
(ii) (c) : Since, D.R.'s are proportional to D.C.'s, therefore L, will be parallel to L₂, iff
\(\frac{l_{1}}{l_{2}}=\frac{m_{1}}{m_{2}}=\frac{n_{1}}{n_{2}}\)
(iii) (c) : Equation of line joining Band C is \(\frac{x-1}{4}=\frac{y-4}{0}=\frac{z-6}{-2}\)
Let coordinates of foot of perpendicular be D(x, y, z).
ஃD.R.'s of AD are < x-1, y-2, z-1 >.
Now, 4(x - 1) + 0(y - 2) -2(z - 1) = 0 ⇒ 4x - 2z = 2
Also, (x, y, z) will satisfy equation of line BC.
Here, (3, 4, 5) satisfy both the conditions.
ஃ  Required coordinates are (3, 4, 5).
(iv) (b) : Let a, b, c be the direction ratios of the required line. Since it is perpendicular to the lines
whose direction ratios are (1, -2, -2) and (0, 2, 1) respectively.
ஃ  a - 2b - 2c = 0               ..... (i)
    0 .a + 2b + c = 0           .......(ii)
On solving (i) and (ii) by cross-multiplication, we get
\(\frac{a}{-2+4}=\frac{b}{0-1}=\frac{c}{2} \Rightarrow \frac{a}{2}=\frac{b}{-1}=\frac{c}{2}\)
Thus, the direction ratios of the required line are  < 2, -1, 2 >.
(v) (b) : D.R 's of given lines are < 3, -2, 0 > and < 1, -\(\frac{3}{2}\) ,2 >
Now, as 3.1 + (-2).(\(\frac{3}{2}\)) +0·2 = 3 -3 + 0 = 0
ஃ Given lines are perpendicular to each other.

(i) (b) : The equation of plane passing through three non - collinear points is given by
\(\left|\begin{array}{ccc} x-x_{1} & y-y_{1} & z-z_{1} \\ x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \end{array}\right|=0\)
\(\Rightarrow\left|\begin{array}{ccc} x & y-1 & z-2 \\ 3-0 & 4-1 & -1-2 \\ 2-0 & 4-1 & 2-2 \end{array}\right|=0\)
\(\Rightarrow\left|\begin{array}{ccc} x & y-1 & z-2 \\ 3 & 3 & -3 \\ 2 & 3 & 0 \end{array}\right|=0\)
\( \Rightarrow \ln x(0+9)-(y-1)(0+6)+(z-2)(9-6)=0 \)
\(\Rightarrow 9 x-6 y+6+3 z-6=0 \Rightarrow 3 x-2 y+z=0\)
(ii) (c) : Height of tower = Perpendicular distance from the point (6, 5, 9) to the plane 3x - 2y + z = 0
\(=\left|\frac{18-10+9}{\sqrt{3^{2}+(-2)^{2}+1^{2}}}\right|=\frac{17}{\sqrt{14}} \text { units }\)
(iii) (b) : D.R.'s of perpendicular are < 3, -2, 1 >
[ஃ Perpendicular is parallel to the normal to the plane] Since, perpendicular is passing through the point (6, 5, 9), therefore its equation is 
\(\frac{x-6}{3}=\frac{y-5}{-2}=\frac{z-9}{1}\)
(iv) (a) : Let the coordinates of foot of perpendicular  are
Since, this point lie on the plane 3x - 2y + z = 0, therefore we get, \((3 \lambda+6,-2 \lambda+5, \lambda+9)\)
\( 3(3 \lambda+6)-2(-2 \lambda+5)+(\lambda+9)=0 \)
\(\Rightarrow 9 \lambda+4 \lambda+\lambda+18-10+9=0 \)
\(\Rightarrow 14 \lambda=-17 \Rightarrow \lambda=\frac{-17}{14}\)
Thus, the coordinates of foot of perpendicular are
\(\left(\frac{-51}{14}+6, \frac{34}{14}+5, \frac{-17}{14}+9\right) \text { i.e., }\left(\frac{33}{14}, \frac{104}{14}, \frac{109}{14}\right)\)
(v) (b) : Clearly, area of ABC =\(\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{A C}|\)
\( =\frac{1}{2}|(3 \hat{i}+3 \hat{j}-3 \hat{k}) \times(2 \hat{i}+3 \hat{j})| \)
\(=\frac{1}{2}\left\|\begin{array}{cc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 3 & -3 \\ 2 & 3 & 0 \end{array}\right\|=\frac{1}{2}|9 \hat{i}-6 \hat{j}+3 \hat{k}|\)
\( =\frac{1}{2} \sqrt{9^{2}+6^{2}+3^{2}}=\frac{1}{2} \sqrt{126}=\frac{3}{2} \sqrt{14} \text { sq. units } \)