(i) (b): Let B(x, y) be the point of intersection of the given lines
2x + 5y = 100 ....(i)
and \(\frac{x}{25}+\frac{y}{40}=1 \Rightarrow 8 x+5 y=20\)...(ii)
Solving (i) and (ii), we get
\(x=\frac{50}{3}, y=\frac{40}{3}\)
ஃ The point of intersection \(B(x, y)=\left(\frac{50}{3}, \frac{40}{3}\right)\)
(ii) (d): The corner points of the feasible region shown in the given graph are
\((0,0), A(25,0), B\left(\frac{50}{3}, \frac{40}{3}\right), C(0,20)\)
(iii) (a): Here Z = x + y

Thus, max Z = 30 occurs at point \(\left(\frac{50}{3}, \frac{40}{3}\right)\)
(iv) (b): 

(v) (c):

(i) (c)
(ii) (b): Since (8, 12) satisfy all the inequalities therefore (8, .1.2) is the point in its feasible region.
(iii) (c) : At (0, 0), z = 0
At (16, 0), z = 352
At (8, 12), z = 392
At (0, 20), z = 360
It can be observed that max z occur at (8, 12). Thus, z will attain its optimal value at (8, 12).
(iv) (c) : We have, x + y = 20 ... (i)
and 3x + 2y = 48.. (ii)
On solving (i) and (ii), we get
X = 8, y = 12.
Thus, the coordinates of P are (8,12) and hence (8, 12) is one of its corner points.
(v) (b): The optimal solution occurs at every point on the line joining these two points.

(i) (c) : For all values of x, y = x² + 7
\(\therefore\) Arun's position at any point of x will be (x, x²+ 7)
(ii) (c) : Distance between
Arun and Manita, i.e., D
\(=\sqrt{(x-3)^{2}+\left(x^{2}+7-7\right)^{2}}\) 
\(=\sqrt{(x-3)^{2}+x^{4}}\) 
(iii) (a) : We have \(D=\sqrt{(x-3)^{2}+x^{4}}\) 
\(\therefore \quad D^{2}=(x-3)^{2}+x^{4}\) 
Now, \(\frac{d}{d x}\left(D^{2}\right)=2(x-3)+4 x^{3}=0\) 
\(\Rightarrow 4 x^{3}+2 x-6=0 \Rightarrow 2 x^{3}+x-3=0\) 
\(\Rightarrow \quad(x-1)\left(2 x^{2}+2 x+3\right)=0\) 
\(\therefore\) x = 1
(\(\therefore\) 2x² + 2x + 3 = 0 will give imaginary values)
(iv) (d) : We have,\(D=\sqrt{(x-3)^{2}+x^{4}}\) 
\(D^{\prime}(x)=\frac{2(x-3)+4 x^{3}}{2 \sqrt{(x-3)^{2}+x^{4}}}=0\) 
 
\(\Rightarrow 2 x^{3}+x-3=0\) 
\(\Rightarrow x=1\) 
Clearly, D"(x) at x = 1 is > 0
\(\therefore\) Value of x for which Dwill be minimum is 1
For x = 1, y = 8.
Thus, the required position is (1, 8).
(v) (d) : Minimum value of \(D=\sqrt{(1-3)^{2}+(1)^{4}}\) 
\(=\sqrt{4+1}=\sqrt{5}\)

(i) (b) : Given, perimeter of window = 10 m 
\(\therefore\) x + y + y + perimeter of semicircle = 10
\(\Rightarrow x+2 y+\pi \frac{x}{2}=10\) 
(ii) (b) : \(A=x \cdot y+\frac{1}{2} \pi\left(\frac{x}{2}\right)^{2}\) 
\(=x\left(5-\frac{x}{2}-\frac{\pi x}{4}\right)+\frac{1}{2} \frac{\pi x^{2}}{4}\left[\because \text { From }(\mathrm{i}), y=5-\frac{x}{2}-\frac{\pi x}{4}\right]\) 
\(=5 x-\frac{x^{2}}{2}-\frac{\pi x^{2}}{4}+\frac{\pi x^{2}}{8}=5 x-\frac{x^{2}}{2}-\frac{\pi x^{2}}{8}\) 
(iii) (c) : We have, \(A=5 x-\frac{x^{2}}{2}-\frac{\pi x^{2}}{8}\) 
\(\Rightarrow \quad \frac{d A}{d x}=5-x-\frac{\pi x}{4}\) 
Now, \(\frac{d A}{d x}=0 \Rightarrow 5=x+\frac{\pi x}{4}\) 
\(\Rightarrow x(4+\pi)=20 \Rightarrow x=\frac{20}{4+\pi}\) 
\(\left[\text { Clearly, } \frac{d^{2} A}{d x^{2}}<0 \text { at } x=\frac{20}{4+\pi}\right]\) 
(iv) (d) : At \(x=\frac{20}{4+\pi}\) 
 \(A=5\left(\frac{20}{4+\pi}\right)-\left(\frac{20}{4+\pi}\right)^{2} \frac{1}{2}-\frac{\pi}{8}\left(\frac{20}{4+\pi}\right)^{2}\) 
\(=\frac{100}{4+\pi}-\frac{200}{(4+\pi)^{2}}-\frac{50 \pi}{(4+\pi)^{2}}\) 
\(=\frac{(4+\pi)(100)-200-50 \pi}{(4+\pi)^{2}}=\frac{400+100 \pi-200-50 \pi}{(4+\pi)^{2}}\) 
\(=\frac{200+50 \pi}{(4+\pi)^{2}}=\frac{50(4+\pi)}{(4+\pi)^{2}}=\frac{50}{4+\pi}\) 
(v) (a) : We have, \(y=5-\frac{x}{2}-\frac{\pi x}{4}=5-x\left(\frac{1}{2}+\frac{\pi}{4}\right)\) 
\(=5-x\left(\frac{2+\pi}{4}\right)=5-\left(\frac{20}{4+\pi}\right)\left(\frac{2+\pi}{4}\right)\) 
\(=5-5 \frac{(2+\pi)}{4+\pi}=\frac{20+5 \pi-10-5 \pi}{4+\pi}=\frac{10}{4+\pi}\)

Let A be the event that the student is selected for LIC AAO, B be the event that the student is selected for SSC CGL and C be the event that the student is selected for Bank P.O.
Then, P(A) = a; P(B) = band P(C) = c
(i) (b): We have, \(P(A \cup B \cup C)=0.5\)
\(\Rightarrow \ 1-P(\overline{A \cup B \cup C})=0.5 \)
\(\Rightarrow 1-P(\bar{A} \cap \bar{B} \cap \bar{C})=0.5 \)
\(\Rightarrow 1-P(\bar{A}) \cdot P(\bar{B}) \cdot P(\bar{C})=0.5 \)
\(\Rightarrow\) 1 - (I - a) (1 - b) (1 - c) = 0.5
\(\Rightarrow\) 1 - [ (I - a - b + ab )(1 - c)] = 0.5
\(\Rightarrow\) 1 - [1 - c - a + ac - b + bc + ab - abc] = 0.5
\(\Rightarrow\) a + b + c - ab - bc - ca + abc = 0.5 .......(i)
(ii) (c): We have, P(selection in at least two competitive exams) = 0.4
\( \Rightarrow \ P(A \cap B \cap \bar{C})+P(\bar{A} \cap B \cap C)+P(A \cap \bar{B} \cap C)+ P(A \cap B \cap C)=0.4 \)
\(\Rightarrow\) ab(1 - c) + (1 - a)bc + a(1 - b)c + abc = 0.4
\(\Rightarrow\) ab - abc + bc - abc + ac - abc + abc = 0.4
\(\Rightarrow\) ab + bc + ac - 2abc= 0.4 ......................(ii)
(iii) (a) : We have, P(selection in exactly two examinations) = 0.3
\(\Rightarrow P(A \cap B \cap \bar{C})+P(\bar{A} \cap B \cap C)+P(A \cap \bar{B} \cap C)=0.3\)
\(\Rightarrow\) ab(1 - c) + (1 - a)bc + a(1 - b)c = 0.3
\(\Rightarrow\) ab + bc + ac - 3abc = 0.3 ...................(iii)
On subtracting (iii) from (ii), we get
abc = 0.1
(iv) (b): On substituting the value of abe in (ii), we get
ab + bc + ac = 0.6
(v) (a): On substituting the value of ab + bc + ac and
abc in (i), we get
a + b + c - 0.6 + 0.1 = 0.5
\(\Rightarrow\) a + b + c = 1

Here, \(\left|\vec{F}_{1}\right|=\sqrt{(4)^{2}+0^{2}}=4 \mathrm{KN}\)
\( \left|\vec{F}_{2}\right|=\sqrt{(-2)^{2}+4^{2}}=\sqrt{20} \mathrm{KN} \)
\(\left|\vec{F}_{3}\right|=\sqrt{(-3)^{2}+(-3)^{2}}=\sqrt{18} \mathrm{KN}\)
(i) (a): Since, \(\sqrt 20\) is larger. So, team B will win the game.
(ii) (b): Let F be the combined force
\(\therefore \vec{F}=\vec{F}_{1}+\vec{F}_{2}+\vec{F}_{3}=4 \hat{i}+0 \hat{j}-3 \hat{i}-3 \hat{j}-2 \hat{i}+4 \hat{j} \)
\(=-\hat{i}+\hat{j} \)
\(\therefore |\vec{F}|=\sqrt{(-1)^{2}+1^{2}}=\sqrt{2}=1.4 \mathrm{KN}\)
(iii) (c) : We have,  \(\vec{F}=-\hat{i}+\hat{j}\)
\(\therefore \theta=\tan ^{-1}\left(\frac{F_{y}}{F_{x}}\right)=\tan ^{-1}\left(\frac{1}{-1}\right)=\frac{3 \pi}{4} \text { radian }\)
= 0.75 x 3.14 radian = 2.3555 radian ≈ 2.4 radian 
(iv) (a): Magnitude of force of Team B = \(\sqrt 20\) KN
= 2\(\sqrt 5\) KN
(v) (b): 4 KN force is applied by team A.

(i) (b): P(X = 2) = (Probability of not getting six at first throw) \(\times\) (Probability of getting six at second throw)
\(=\frac{5}{6} \times \frac{1}{6}=\frac{5}{6^{2}}\)
(ii) (c): P(X =4) = \(\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}=\frac{5^{3}}{6^{4}}\)
(iii) (c) : P(X ≥  2) = 1 - P(X < 2)
\(=1-P(X=1)=1-\frac{1}{6}=\frac{5}{6}\)
(iv) (a): \(P(X \geq 6)=\left(\frac{5}{6}\right)^{5} \times \frac{1}{6}+\left(\frac{5}{6}\right)^{6} \times \frac{1}{6}+\ldots \infty\)
\(=\frac{5^{5}}{6^{6}}\left[1+\frac{5}{6}+\left(\frac{5}{6}\right)^{2}+\ldots \infty\right]=\frac{5^{5}}{6^{6}}\left[\frac{1}{1-\frac{5}{6}}\right]=\left(\frac{5}{6}\right)^{5}\)
(v) (d): \(P(X \geq 4)=\left(\frac{5}{6}\right) \cdot \frac{1}{6}+\left(\frac{5}{6}\right)^{4} \cdot \frac{1}{6}+\ldots \ldots\)
\( =\frac{5^{3}}{6^{4}}\left[1+\left(\frac{5}{6}\right)+\left(\frac{5}{6}\right)^{2}+\ldots . .\right] \)
\(=\frac{5^{3}}{6^{4}}\left[\frac{1}{1-\left(\frac{5}{6}\right)}\right]=\left(\frac{5^{3}}{6^{4}}\right) \cdot 6=\frac{5^{3}}{6^{3}}\)

(i) (b): Here (5, 3) are the coordinates of B.
\(\therefore\) P.V. of B = \(5 \hat{i}+3 \hat{j}\)
(ii) (d) : Here (9, 8) are the coordinates of D.
\(\therefore\) P.V. of D = \({9} \hat{i}+8 \hat{j}\)
(iii) (b) : P.V. of B =  \(5 \hat{i}+3 \hat{j}\)     and ஃ P.V. of C = \(6 \hat{i}+5 \hat{j}\)
\(\therefore \ \overrightarrow{B C}=(6-5) \hat{i}+(5-3) \hat{j}=\hat{i}+2 \hat{j}\)
(iv) (b): Since P.V. of A = \(2 \hat{i}+2 \hat{j}\) , P.V. of D = \({9} \hat{i}+8 \hat{j}\)
\( \therefore \overrightarrow{A D}=(9-2) \hat{i}+(8-2) \hat{j}=7 \hat{i}+6 \hat{j} \)
\(|\overrightarrow{A D}|^{2}=7^{2}+6^{2}=49+36=85 \)
\(\Rightarrow |\overrightarrow{A D}|=\sqrt{85} \text { units }\)
(v) (a): We have, \(\vec{M}=4 \hat{j}+3 \hat{k}\) ,
\( \therefore |\vec{M}|=\sqrt{4^{2}+3^{2}}=\sqrt{16+9}=\sqrt{25}=5 \)
\(\therefore \hat{M}=\frac{\vec{M}}{|\vec{M}|}=\frac{4 \hat{j}+3 \hat{k}}{5}=\frac{4}{5} \hat{j}+\frac{3}{5} \hat{k}\)

(i) (d) : Line x + y = 3 cuts the x-axis and y-axis at (3, 0) and (0, 3) respectively
[Since, at x-axis, y = 0 and at y-axis, x = 0]
(ii) (c) : We have, y² = 4x and x + y = 3
From (i) and (ii), we have y² =  4(3 - y)
\(\Rightarrow y^{2}+4 y-12=0 \Rightarrow y^{2}+6 y-2 y-12=0\) 
\(\Rightarrow y(y+6)-2(y+6)=0\) 
\(\Rightarrow (y+6)(y-2)=0 \Rightarrow y=2, y=-6\) 
From (ii), x = 3 - 2 = 1 or x = 3 + 6 = 9
\(\therefore\) Required points of intersection are (1, 2), (9, - 6)
(iii) (b) :

(iv) (d): \(\int_{-6}^{2}(3-y) d y=\left[3 y-\frac{y^{2}}{2}\right]_{-6}^{2}\) 
\(=\left[6-\frac{4}{2}-\left[3(-6)-\frac{(-6)^{2}}{2}\right]\right]=4+36=40\) 
(v) (c): Required area = \(\int_{-6}^{2}(3-y) d y-\int_{-6}^{2} \frac{y^{2}}{4} d y\) 
\(=40-\frac{1}{4}\left[\frac{y^{3}}{3}\right]_{-6}^{2}=40-\frac{1}{4}\left[\frac{8}{3}-\frac{(-6)^{3}}{3}\right]\) 
\(=40-\frac{2}{3}-\frac{216}{12}=\frac{480-8-216}{12}=\frac{256}{12}=\frac{64}{3} \mathrm{sq} . \text { units }\)

(i) (a) : Equation of line AB is
\(y-0=\frac{3-0}{1+1}(x+1) \Rightarrow y=\frac{3}{2}(x+1)\) 
(ii) (c) : Equation of line BC is \(y-3=\frac{2-3}{3-1}(x-1)\) 
\(\Rightarrow y=-\frac{1}{2} x+\frac{1}{2}+3 \Rightarrow y=\frac{-1}{2} x+\frac{7}{2}\) 
(iii) (d) : Area of region ABCD
= Area of \(\triangle A B E\) + Area of region BCDE
\(=\int_{-1}^{1} \frac{3}{2}(x+1) d x+\int_{1}^{3}\left(\frac{-1}{2} x+\frac{7}{2}\right) d x\) 
\(=\frac{3}{2}\left[\frac{x^{2}}{2}+x\right]_{-1}^{1}+\left[\frac{-x^{2}}{4}+\frac{7}{2} x\right]_{1}^{3}\) 
\(=\frac{3}{2}\left[\frac{1}{2}+1-\frac{1}{2}+1\right]+\left[\frac{-9}{4}+\frac{21}{2}+\frac{1}{4}-\frac{7}{2}\right]\) 
= 3 + 5 = 8 sq. units
(iv) (a) : Equation of line AC is \(y-0=\frac{2-0}{3+1}(x+1)\) 
\(\Rightarrow y=\frac{1}{2}(x+1)\) 
\(\therefore \text { Area of } \Delta A D C=\int_{-1}^{3} \frac{1}{2}(x+1) d x=\left[\frac{x^{2}}{4}+\frac{1}{2} x\right]_{-1}^{3}\) 
\(=\frac{9}{4}+\frac{3}{2}-\frac{1}{4}+\frac{1}{2}=4 \text { sq. units }\) 
(v) (b) : Area of \(\Delta A B C\)= Area of region ABCD - Area of \(\Delta A C D=8-4=4 \mathrm{sq} . \text { units }\)

(i) (a) : \(\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}, \vec{b}=3 \hat{i}-3 \hat{j}-2 \hat{k}\)  and \(\vec{c}=-2 \hat{i}+2 \hat{j}+6 \hat{k}\)
\(\therefore \ \vec{a}+\vec{b}+\vec{c}=2 \hat{i}+3 \hat{j}+6 \hat{k}\)
(ii) (c): Using triangle law of addition in \(\Delta\)ABC, we get
\(\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C A}=\overrightarrow{0}\) which can be rewritten as
\(\overrightarrow{A B}+\overrightarrow{B C}-\overrightarrow{A C}=\overrightarrow{0} \text { or } \overrightarrow{A B}-\overrightarrow{C B}+\overrightarrow{C A}=\overrightarrow{0}\)
(iii) (c) : We have, A(1, 4, 2), B(3, -3, -2) and C(-2, 2, 6)
Now, \(\overrightarrow{A B}=\vec{b}-\vec{a}=2 \hat{i}-7 \hat{j}-4 \hat{k}\)
and \(\overrightarrow{A C}=\vec{c}-\vec{a}=-3 \hat{i}-2 \hat{j}+4 \hat{k}\)
\(\therefore \overrightarrow{A B} \times \overrightarrow{A C}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & -7 & -4 \\ -3 & -2 & 4 \end{array}\right|\)
\(=\hat{i}(-28-8)-\hat{j}(8-12)+\hat{k}(-4-21)=-36 \hat{i}+4 \hat{j}-25 \hat{k}\)
Now, \(|\overrightarrow{A B} \times \overrightarrow{A C}|=\sqrt{(-36)^{2}+4^{2}+(-25)^{2}}\)
\(=\sqrt{1296+16+625}=\sqrt{1937}\)
\(\therefore \text { Area of } \Delta A B C=\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{A C}|=\frac{1}{2} \sqrt{1937} \text { sq. units }\)
(iv) (d): If the given points lie on the straight line, then the points will be collinear and so area of \( \Delta A B C=0 \)
\(\Rightarrow |\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}|=0\)
If a, b, c are the position vectors of the three vertices A, Band C of \( \Delta A B C \), then area of triangle
\(\left.=\frac{1}{2}|\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}|\right]\)
(v) (b): Here, \( |\vec{a}| =\sqrt{2^{2}+3^{2}+6^{2}}=\sqrt{4+9+36} \)
\(=\sqrt{49}=7\)
\(\therefore\) Unit vector in the direction of vector \(\vec{a}\) is
\(\hat{a}=\frac{2 \hat{i}+3 \hat{j}+6 \hat{k}}{7}=\frac{2}{7} \hat{i}+\frac{3}{7} \hat{j}+\frac{6}{7} \hat{k}\)

(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 : Curves y = cos x and y = x + 1 meet at point C(O, 1).

(ii) C : curve y=cosx meet the x axis at \(A^{\prime}\left(\frac{-\pi}{2}, 0\right)\) and \(A\left(\frac{\pi}{2}, 0\right)\) .
(iii) (a) : \(\int_{-1}^{0}(x+1) d x=\left[\frac{x^{2}}{2}+x\right]_{-1}^{0}=0-\left(\frac{1}{2}-1\right)=\frac{1}{2}\) 
(iv) (d) : \(\int_{0}^{\pi / 2} \cos x d x=[\sin x]_{0}^{\pi / 2}=\sin \frac{\pi}{2}-\sin 0=1\) 
(v) (b) : Required area \(\int_{-1}^{0}(x+1) d x+\int_{0}^{\pi / 2} \cos x d x\) 
\(=\frac{1}{2}+1=\frac{3}{2} \text { sq. units }\)

(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) (d) :Given, at any time t the thermometer reading be T^oF and the outside temperature be sop.
Then, by Newton's law of cooling, we have
\(\frac{d T}{d t} \propto(T-S) \Rightarrow \frac{d T}{d t}=-\lambda(T-S)\) 
(ii) (d) : Since, after 5 minutes, thermometer reads 60⁰F.
\(\therefore\) Value of T(5) = 60^0F
(iii) (a) : Clearly from given information, value of T( 10) is 50°F.
(iv) (b) : We have,\(\frac{d T}{d t}=-\lambda(T-S)\) 
\(\Rightarrow \frac{d T}{T-S}=-\lambda d t \Rightarrow \int \frac{1}{T-S} d T=-\lambda \int d t\) 
\(\Rightarrow \log (T-S)=-\lambda t+c\) 
(v) (c) : Since, at t = 0, T = 80°