To solve the problem, we need to find the points of intersection of the given line with the xy-plane and yz-plane, and then calculate the volume of the tetrahedron formed by the origin and these points along with point C.
### Step 1: Find the intersection with the xy-plane
The equation of the line is given as:
\[
\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-4}{4}
\]
To find the intersection with the xy-plane, we set \( z = 0 \).
Substituting \( z = 0 \) into the line equation:
\[
\frac{x-1}{2} = \frac{y-2}{3} = \frac{0-4}{4} = -1
\]
Now we can solve for \( x \) and \( y \):
1. From \( \frac{x-1}{2} = -1 \):
\[
x - 1 = -2 \implies x = -1
\]
2. From \( \frac{y-2}{3} = -1 \):
\[
y - 2 = -3 \implies y = -1
\]
Thus, the coordinates of point A (intersection with the xy-plane) are:
\[
A(-1, -1, 0)
\]
### Step 2: Find the intersection with the yz-plane
To find the intersection with the yz-plane, we set \( x = 0 \).
Substituting \( x = 0 \) into the line equation:
\[
\frac{0-1}{2} = \frac{y-2}{3} = \frac{z-4}{4}
\]
This gives:
\[
\frac{-1}{2} = \frac{y-2}{3} \quad \text{and} \quad \frac{-1}{2} = \frac{z-4}{4}
\]
Now we can solve for \( y \) and \( z \):
1. From \( \frac{y-2}{3} = -\frac{1}{2} \):
\[
y - 2 = -\frac{3}{2} \implies y = \frac{1}{2}
\]
2. From \( \frac{z-4}{4} = -\frac{1}{2} \):
\[
z - 4 = -2 \implies z = 2
\]
Thus, the coordinates of point B (intersection with the yz-plane) are:
\[
B(0, \frac{1}{2}, 2)
\]
### Step 3: Identify point C
Point C is given as:
\[
C(1, 0, 4)
\]
### Step 4: Calculate the volume of tetrahedron OABC
The volume \( V \) of the tetrahedron formed by points O (origin), A, B, and C can be calculated using the formula:
\[
V = \frac{1}{6} \left| \vec{OA} \cdot (\vec{OB} \times \vec{OC}) \right|
\]
#### Step 4.1: Find vectors OA, OB, and OC
- \( \vec{OA} = A - O = (-1, -1, 0) \)
- \( \vec{OB} = B - O = (0, \frac{1}{2}, 2) \)
- \( \vec{OC} = C - O = (1, 0, 4) \)
#### Step 4.2: Calculate the cross product \( \vec{OB} \times \vec{OC} \)
\[
\vec{OB} \times \vec{OC} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
0 & \frac{1}{2} & 2 \\
1 & 0 & 4
\end{vmatrix}
\]
Calculating the determinant:
\[
= \hat{i} \left( \frac{1}{2} \cdot 4 - 2 \cdot 0 \right) - \hat{j} \left( 0 \cdot 4 - 2 \cdot 1 \right) + \hat{k} \left( 0 \cdot 0 - \frac{1}{2} \cdot 1 \right)
\]
\[
= \hat{i} \cdot 2 + \hat{j} \cdot 2 - \hat{k} \cdot \frac{1}{2}
\]
Thus,
\[
\vec{OB} \times \vec{OC} = (2, 2, -\frac{1}{2})
\]
#### Step 4.3: Calculate the dot product \( \vec{OA} \cdot (\vec{OB} \times \vec{OC}) \)
\[
\vec{OA} \cdot (\vec{OB} \times \vec{OC}) = (-1, -1, 0) \cdot (2, 2, -\frac{1}{2})
\]
\[
= -1 \cdot 2 + -1 \cdot 2 + 0 \cdot -\frac{1}{2} = -2 - 2 + 0 = -4
\]
#### Step 4.4: Calculate the volume
\[
V = \frac{1}{6} \left| -4 \right| = \frac{4}{6} = \frac{2}{3}
\]
### Step 5: Calculate \( 102V \)
\[
102V = 102 \cdot \frac{2}{3} = 68
\]
Thus, the value of \( 102V \) is:
\[
\boxed{68}
\]
