To solve this problem, we need to find a line $L$ that is orthogonal to the two given lines and passes through the origin.

Step 1: Identify the Direction Vectors

For the lines given in the problem:

1. First line: $x = 4 + 4t, \quad y = 6 + 3t, \quad z = 7 - t$ The direction vector of this line, which we will call $\vec{d}_1$, can be found from the coefficients of $t$: $\vec{d}_1 = \langle 4, 3, -1 \rangle$

2. Second line: $x = 2 - 7s, \quad y = -1 + 5s, \quad z = 3 + 2s$ The direction vector of this line, $\vec{d}_2$, can be found from the coefficients of $s$: $\vec{d}_2 = \langle -7, 5, 2 \rangle$

Step 2: Find a Vector Orthogonal to Both Lines

A vector that is orthogonal to both $\vec{d}_1$ and $\vec{d}_2$ can be found by taking the cross product $\vec{d}_1 \times \vec{d}_2$.