Since, $$T=\left\{x \left\lvert\, \frac{x+5}{x-7}-5=\frac{4 x-40}{13-x}\right.\right\}$$

$\begin{array}{ll}\because & \frac{x+5}{x-7}-5=\frac{4 x-40}{13-x} \\ \Rightarrow & \frac{x+5-5(x-7)}{x-7}=\frac{4 x-40}{13-x} \\ \Rightarrow & \frac{x+5-5 x+35}{x-7}=\frac{4 x-40}{13-x} \\ \Rightarrow & \frac{-4 x+40}{x-7}=\frac{4 x-40}{13-x} \\ \Rightarrow & -(4 x-40)(13-x)=(4 x-40)(x-7)\end{array}$

$\Rightarrow \quad(4 x-40)(x-7)+(4 x-40)(13-x)=0$

$$\begin{array}{rlr} \Rightarrow & (4 x-40)(x-7+13-x) =0 \\ \Rightarrow & 4(x-10) 6 =0 \\ \Rightarrow & 24(x-10) =0 \\ \Rightarrow & x =10 \\ \therefore & T =\{10\} \end{array}$$

Hence, T is not an empty set.

$$\begin{array}{ll} \text { Let } & x \in A \cap(B \cup C) \\ \Rightarrow & x \in A \text { and } x \in(B \cup C) \\ \Rightarrow & x \in A \text { and }(x \in B \text { or } x \in C) \\ \Rightarrow & (x \in A \text { and } x \in B) \text { or }(x \in A \text { and } x \in C) \\ \Rightarrow & x \in A \cap B \text { or } x \in A \cap C \\ & x \in(A \cap B) \cup(A \cap C) \\ \Rightarrow & A \cap(B \cup C) \subset(A \cap B) \cup(A \cap C)\quad \text{... (i)} \end{array}$$

$\begin{array}{ll}\text { Again, let } & y \in(A \cap B) \cup(A \cap C) \\ \Rightarrow & y \in(A \cap B) \text { or } y \in(A \cap C) \\ \Rightarrow & (y \in A \text { and } y \in B) \text { or }(y \in A \text { and } y \in C) \\ \Rightarrow & y \in A \text { and }(y \in B \text { or } y \in C) \\ \Rightarrow & y \in A \text { and } y \in B \cup C \\ \Rightarrow & y \in A \cap(B \cup C) \\ \Rightarrow & (A \cap B) \cup(A \cap C) \subset A \cap(B \cup C)\quad \text{.... (ii)}\end{array}$

From Eqs.(i) and( ii),

$$A \cap(B \cup C)=(A \cap B) \cup(A \cap C)$$

Let $M$ be the set of students who passed in Mathematics, $E$ be the set of students who passed in English and $S$ be the set of students who passed in Science.

Then,

$$\begin{aligned} n(u) & =100 \\ n(E)=15, n(M) & =12, n(S)=8, n(E \cap M)=6, n(M \cap S)=7, \\ n(E \cap S) & =4, \text { and } n(E \cap M \cap S)=4 \\ \because \quad n(E) & =15 \end{aligned}$$

$$\begin{array}{lr} \Rightarrow & a+b+e+f=15 \quad \text{.... (i)}\\ \text { and } & n(M)=12 \\ \Rightarrow & b+c+e+d=12 \quad \text{.... (ii)} \end{array}$$

Also, $$n(S)=8$$

$\Rightarrow \quad d+e+f+g=8$ ... (iii)

$n(E \cap M)=6$

$\Rightarrow \quad b+e=6$ .... (iv)

$n(M \cap S)=7$

$\Rightarrow \quad e+d=7$ .... (v)

$n(E \cap S)=4$

$\Rightarrow \quad e+f=4$ .... (vi)

$n(E \cap M \cap S)=4$

$\Rightarrow \quad e=4$ .... (vii)

From Eqs. (vi) and (vii), $$f=0$$

From Eqs. (v) and (vii), $d=3$

From Eqs. (iv) and (vii), $b=2$

On substituting the values of $d, e$ and $f$ in Eq. (iii), we get

$$\begin{aligned} 3+4+0+g & =8 \\ g & =1 \end{aligned}$$

On substituting the value of $b, e$ and $d$ in Eq. (ii), we get

$$\begin{array}{rlrl} & & 2+c+4+3 & =12 \\ \Rightarrow & c & =3 \end{array}$$

On substituting b, $e$, and $f$ in Eq. (i), we get

$$\begin{aligned} a+2+4+0 & =15 \\ a & =9 \end{aligned}$$

Alternate Method Let $E$ denotes the set of student who passed in English. $M$ denotes the set of students who passed in Mathematics. $S$ denotes the set of students who passed in Science.

Now,

$$\begin{aligned} n(U) & =100, n(E)=15, n(m)=12, n(S)=8 \\ n(E \cap M) & =6, n(M \cap S)=7 \\ n(E \cap S) & =4, n(E \cap M \cap S)=4 \end{aligned}$$

(i) Number of students passed in English and Mathematics but not in Science i.e.,

$$\begin{aligned} n\left(E \cap M \cap S^{\prime}\right) & =n(E \cap M)-n(E \cap M \cap S) \quad \left[\because A \cap B^{\prime}=A-(A \cap B)\right] \\ & =6-4=2 \end{aligned}$$

(ii) Number of students passed in Mathematics and Science but not in English.

$$\begin{aligned} \text { i.e., } \quad n\left(M \cap S \cap E^{\prime}\right) & =n(M \cap S)-n(M \cap S \cap E) \\ & =7-4=3 \end{aligned}$$

(iii) Number of students passed in mathematics only

$$\begin{aligned} \text { i.e., } \quad n\left(M \cap S^{\prime} \cap E^{\prime}\right) & =n(M)-n(M \cap S)-n(M \cap E)+n(M \cap S \cap E) \\ & =12-7-6+4=3 \end{aligned}$$

(iv) Number of students passed in more than one subject only

$$\begin{aligned} \text { i.e., } \quad n(E \cap M)+n(M & \cap S)+n(E \cap S)-3 n(E \cap M \cap S)+n(E \cap M \cap S) \\ & =6+7+4-4 \times 3+4 \\ & =17-12+4=5+4=9 \end{aligned}$$

Let $C$ be the set of students who play cricket and $T$ be the set of students who play tennis.

Then, $$n(U)=60, n(C)=25, n(T)=20, \text { and } n(C \cap T)=10$$

$$\therefore \quad n(C \cup T)=n(C)+n(T)-n(C \cap T)$$

$=25+20-10=35$

$$\begin{aligned} \therefore \text { Number of students who play neither } & =n(U)-n(C \cup T) \\ & =60-35=25 \end{aligned}$$

Let $M$ be the set of students who study Mathematics, $P$ be the set of students who study Physics and $C$ be the set of students who study Chemistry.

Then,

$$\begin{aligned} n(U) & =200, n(M)=120, n(P)=90 \\ n(C) & =70, n(M \cap P)=40, n(P \cap C)=30 \\ n(C \cap M) & =50, n\left(M^{\prime} \cap P^{\prime} \cap C^{\prime}\right)=20 \end{aligned}$$

$$\begin{aligned} & & n(U)-n(M \cup P \cup C) & =20, \\ & & n(M \cup P \cup C) & =200-20=180 \\ \because & & n(M \cup P \cup C) & =n(M)+n(P)+n(C)-n(C \cap M)+n(M \cap P \cap C) \\ \Rightarrow & & 180 & =120+90+70-40-30-50+n(M \cap P \cap C) \\ \Rightarrow & & 180 & =160+n(M \cap P \cap C) \\ \Rightarrow & & n(M \cap P \cap C) & =180-160=20 \end{aligned}$$

So, the number of students who study all the three subjects is 20.