Consider the given points $A(5,2), B(2,3)$ and $C(3,-1)$.

Slope of the line passing through the points $B$ and $C, m_{B C}=\frac{-1-3}{3-2}=-4$

So, the slope of required line is $\frac{1}{4}$.

Since, the equation of a line passing the point $A(5,2)$ and having slope $\frac{1}{4}$ is $y-2=\frac{1}{4}(x-5)$.

$$\begin{array}{lr} \Rightarrow & 4 y-8=x-5 \\ \Rightarrow & x-4 y+3=0 \end{array}$$

and $$\begin{aligned} \text{Given lines,}\quad y & =(2-\sqrt{3})(x+5)\quad \text{.... (i)} \\ \text{Slope of this line,}\quad m_1 & =(2-\sqrt{3}) \\ \text{and}\quad y & =(2+\sqrt{3})(x-7) \quad \text{.... (ii)}\\ \text{Slope of this line,}\quad m_2 & =(2+\sqrt{3}) \end{aligned}$$

Let $\theta$ be the angle between lines (i) and (ii), then

$$\begin{array}{ll} & \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right| \\ \Rightarrow & \tan \theta=\left|\frac{(2-\sqrt{3})-(2+\sqrt{3})}{1+(2-\sqrt{3})(2+\sqrt{3})}\right| \Rightarrow \tan \theta=\left|\frac{-2 \sqrt{3}}{1+4-3}\right| \\ \Rightarrow & \tan \theta=\sqrt{3} \\ \Rightarrow & \tan \theta=\tan \pi / 3 \\ \therefore & \theta=\pi / 3=60 \Upsilon \end{array}$$

For obtuse angle $=\pi-\pi / 3=2 \pi / 3=120 \Upsilon$

Hence, the angle between the lines are $60^{\circ}$ or $120^{\circ}$.

Let the intercept along the axes be $a$ and $b$.

Given, $$a+b=14 \Rightarrow b=14-a$$

Now, the equation of line is $\frac{x}{a}+\frac{y}{b}=1\quad \text{.... (i)}$.

$$\Rightarrow \quad \frac{x}{a}+\frac{y}{14-a}=1$$

Since, the point $(3,4)$ lies on the line.

$$\begin{aligned} & \therefore \quad \frac{3}{a}+\frac{4}{14-a}=1 \\ & \Rightarrow \quad \frac{42-3 a+4 a}{a(14-a)}=1 \Rightarrow 42+a=14 a-a^2 \\ & \Rightarrow \quad a^2-13 a+42=0 \Rightarrow a^2-7 a-6 a+42=0 \\ & \Rightarrow \quad a(a-7)-6(a-7)=0 \Rightarrow(a-7)(a-6)=0 \\ & \Rightarrow \quad a-7=0 \text { or } a-6=0 \\ & \therefore \quad a=7 \text { or } a=6 \\ & \text { When } \\ & a=7 \text {, then } b=7 \\ & \text { When } \\ & a=6 \text {, then } b=8 \end{aligned}$$

$\therefore$ The equation of line, when $a=7$ and $b=7$ is

$$\frac{x}{7}+\frac{y}{7}=1 \Rightarrow x+y=7$$

So, the equation of line, when $a=6$ and $b=8$ is $\frac{x}{6}+\frac{y}{8}=1$

Let the required point be $(h, k)$ and point $(h, k)$ lies on the line $x+y=4$

i.e., $$h+k=4\quad \text{.... (i)}$$

The distance of the point $(h, k)$ from the line $4 x+3 y=10$ is

$$\begin{aligned} \left|\frac{4 h+3 k-10}{\sqrt{16+9}}\right| & =1 \\ 4 h+3 k-10 & = \pm 5 \\ \text{Taking positive sign,}\quad 4 h+3 k & =15\quad \text{.... (ii)} \end{aligned}$$

From Eq. (i) $h=4-k$ put in Eq. (ii), we get

$$\begin{aligned} 4(4-k)+3 k & =15 \\ \Rightarrow \quad 16-4 k+3 k & =15 \\ \Rightarrow \quad k & =1 \end{aligned}$$

On putting $k=1$ in Eq. (i), we get

$$h+1=4 \Rightarrow h=3$$

So, the point is $(3,1)$.

Taking negative sign,

$$\begin{array}{ll} & 4 h+3 k-10=-5 \\ \Rightarrow & 4(4-k)+3 k=5 \\ \Rightarrow & 16-4 k+3 k=5 \\ \Rightarrow & -k=5-16=-11 \\ \therefore & k=11 \end{array}$$

On putting $k=11$ in Eq. (i), we get

$$h+11=4 \Rightarrow h=-7$$

Hence, the required points are $(3,1)$ and $(-7,11)$.

$$\begin{aligned} &\text { Given equation of lines are }\\ &\begin{array}{lr} & \frac{x}{a}+\frac{y}{b}=1 \quad \text{... (i)}\\ \therefore & \text { Slope, } m_1=-\frac{b}{a} \\ \text { and } & \frac{x}{a}-\frac{y}{b}=1 \quad \text{... (ii)}\\ \therefore & \text { Slope, } m_2=\frac{b}{a} \end{array} \end{aligned}$$

$$\begin{aligned} &\text { Let } \theta \text { be the angle between the given lines, then }\\ &\begin{aligned} & \tan \theta=\left|\frac{m_1-m_2}{1+m_1 m_2}\right| \Rightarrow \tan \theta=\left|\frac{-\frac{b}{a}-\frac{b}{a}}{1+\left(\frac{-b}{a}\right)\left(-\frac{b}{a}\right)}\right| \\ \Rightarrow \quad & \tan \theta=\left|\frac{\frac{-2 b}{a}}{\frac{a^2-b^2}{a^2}}\right| \Rightarrow \tan \theta=\frac{2 a b}{a^2-b^2} \end{aligned} \end{aligned}$$

Hence proved.