You have shared an image containing a problem that involves finding the equation of a line passing through the point of intersection of two given lines and the origin. Here's a breakdown of the work:

Given Lines:

• $L_1: \frac{x-1}{2} = \frac{y+3}{4} = \frac{z-5}{3}$
• $L_2: \frac{x-4}{2} = \frac{y+3}{2} = \frac{z-14}{4}$

The goal is to:

1. Find the intersection of these two lines.
2. Find the equation of the line passing through the origin and the point of intersection of $L_1$ and $L_2$.

Step 1: Find the point of intersection of $L_1$ and $L_2$

The parametric forms of the lines are obtained as:

Line $L_1$:

$x = 1 + 2\lambda, \quad y = -3 + 4\lambda, \quad z = 5 + 3\lambda$ where $\lambda$ is a parameter for $L_1$.

Line $L_2$:

$x = 4 + 2\mu, \quad y = -3 + 2\mu, \quad z = 14 + 4\mu$ where $\mu$ is a parameter for $L_2$.

Since the lines intersect, their coordinates must be equal at the point of intersection: $1 + 2\lambda = 4 + 2\mu, \quad -3 + 4\lambda = -3 + 2\mu, \quad 5 + 3\lambda = 14 + 4\mu$

From these, we can solve for $\lambda$ and $\mu$:

1. From $1 + 2\lambda = 4 + 2\mu$: \lambda - \mu = \frac{3}{2} \tag{1}

2. From $-3 + 4\lambda = -3 + 2\mu$: 2\lambda = \mu \quad \Rightarrow \quad \mu = 2\lambda \tag{2}

3. Substituting $\mu = 2\lambda$ into equation (1): $\lambda - 2\lambda = \frac{3}{2} \quad \Rightarrow \quad \lambda = -\frac{3}{2}$

Now, substituting $\lambda = -\frac{3}{2}$ into $\mu = 2\lambda$: $\mu = 2 \times -\frac{3}{2} = -3$

Thus, the point of intersection can be found by substituting $\lambda = -\frac{3}{2}$ into the parametric equations of $L_1$ (or $L_2$):

$x = 1 + 2\left(-\frac{3}{2}\right) = 1 - 3 = -2$ $y = -3 + 4\left(-\frac{3}{2}\right) = -3 - 6 = -9$ $z = 5 + 3\left(-\frac{3}{2}\right) = 5 - \frac{9}{2} = \frac{10}{2} - \frac{9}{2} = \frac{1}{2}$

So, the point of intersection is $(-2, -9, \frac{1}{2})$.

Step 2: Find the equation of the line passing through the origin and the intersection point

Now that we have the point of intersection $(-2, -9, \frac{1}{2})$, the equation of the line passing through the origin $(0, 0, 0)$ and this point can be written in parametric form as:

$\frac{x}{-2} = \frac{y}{-9} = \frac{z}{\frac{1}{2}} = t$

This is the equation of the line passing through the origin and the point $(-2, -9, \frac{1}{2})$.

Would you like a more detailed breakdown of any part of this solution? Here are some related questions for further exploration:

1. How can we check if two lines are skew or intersecting?
2. Can you derive the parametric form of a line given two points?
3. What if the lines $L_1$ and $L_2$ were parallel? How would the solution change?
4. How can we find the angle between two lines in 3D space?
5. Can we express the line in vector form instead of parametric form?

Tip: When solving for points of intersection in 3D geometry, consistency between parametric equations is key to ensuring the lines truly intersect.