Find the equation of the line …

Since the family of lines passing through the intersection of given lines is

(2x + 3y - 4) k(x - 5y + 7) = 0

This line meets x-axis i.e. y = 0, then

2x - 4 + k(x - 5y + 9) = 0

x = (4 - 7k)/(2 + k), which is the x-intercept.

Therefore, -4 = (4 - 7k)/(2 + k)

k = 4

Putting the value of k in the family equation

(2x + 3y - 4) + 4(x - 5y + 7) = 0

6x - 17y + 24 = 0, which is the required equation

Ans. $2x + 3y - 4 = 0 => 2x +3y = 4 ........\left(1\right)$

$x - 5y = 7 ........\left(2\right)$

First we need to find intersection of the lines, For that solve these equation for x and y 
multiply equation (2) by 2, We get 

$2x - 10y = 14 ......\left(3\right)$

Subtract (1) from (3), we get 
$-13y = 10 => y = \frac{-10}{13}$

Put value of y in (1), we get  $x = \frac{41}{13}$
 
Let the Equation of line in intercept form is 
$\frac{x}{a} + \frac{y}{b} = 1$

as x intecept is -4  and point $\left(\left(\frac{41}{13}\right),\left(\frac{-10}{13}\right)\right)$ satisfies this equation.

putting all Values, we get

$\frac{41}{13\times -4} + \frac{-10}{13b} = 1$

=> $-\frac{41}{52} - 1 = \frac{10}{13b}$

=> $\frac{10}{13b} = \frac{-93}{52}$

=> $b = -\frac{40}{93}$

So the Equation of line ll be 

$\frac{x}{-4} - \frac{93y}{40} = 1$