Step-by-Step Solution

1. Let the first line be $L_1: \frac{x-1}{2}=\frac{y-b}{3}=\frac{z-3}{4} = \lambda$

   Then, any point on $L_1$ can be written as $P_1(2\lambda + 1, 3\lambda + b, 4\lambda + 3)$

2. Let the second line be $L_2: \frac{x-4}{5}=\frac{y-1}{2}=\frac{z}{1} = \mu$

   Then, any point on $L_2$ can be written as $P_2(5\mu + 4, 2\mu + 1, \mu)$

3. For the lines to intersect, the points $P_1$ and $P_2$ must be the same for some values of $\lambda$ and $\mu$.

   Equating the coordinates, we get the following system of equations:

   1. $2\lambda + 1 = 5\mu + 4$ => $2\lambda - 5\mu = 3$ ...(1)
   2. $3\lambda + b = 2\mu + 1$ => $3\lambda - 2\mu = 1 - b$ ...(2)
   3. $4\lambda + 3 = \mu$ => $4\lambda - \mu = -3$ ...(3)

4. From equation (3), we have $\mu = 4\lambda + 3$. Substituting this into equation (1):

   $2\lambda - 5(4\lambda + 3) = 3$

   $2\lambda - 20\lambda - 15 = 3$

   $-18\lambda = 18$

   $\lambda = -1$

5. Substituting $\lambda = -1$ into $\mu = 4\lambda + 3$, we get:

   $\mu = 4(-1) + 3 = -1$

6. Now, substitute $\lambda = -1$ into equation (2):

   $3(-1) - 2(-1) = 1 - b$

   $-3 + 2 = 1 - b$

   $-1 = 1 - b$

   $b = 2$

7. Now that we have $\lambda = -1$ and $\mu = -1$, we can find the point of intersection.

   Using $L_1$: $P_1(2(-1) + 1, 3(-1) + 2, 4(-1) + 3) = P_1(-1, -1, -1)$

   Using $L_2$: $P_2(5(-1) + 4, 2(-1) + 1, -1) = P_2(-1, -1, -1)$