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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 $P_1(2\lambda + 1, 3\lambda + b, 4\lambda + 3)$
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 $P_2(5\mu + 4, 2\mu + 1, \mu)$
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:
From equation (3), we have $\mu = 4\lambda + 3$. Substituting this into equation (1):
$2\lambda - 5(4\lambda + 3) = 3$
$2\lambda - 20\lambda - 15 = 3$
$-18\lambda = 18$
$\lambda = -1$
Substituting $\lambda = -1$ into $\mu = 4\lambda + 3$, we get:
$\mu = 4(-1) + 3 = -1$
Now, substitute $\lambda = -1$ into equation (2):
$3(-1) - 2(-1) = 1 - b$
$-3 + 2 = 1 - b$
$-1 = 1 - b$
$b = 2$
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 $L_2$: $P_2(5(-1) + 4, 2(-1) + 1, -1) = P_2(-1, -1, -1)$
Correct Answer: b = 2, Point of Intersection: (-1, -1, -1)