If the line `(x-1)/(2)=(y-2)/(3)=(z-4)/(4)` intersect the xy and yz plane at points A and B respectively. If the volume of the tetrahedron OABC is V cubic units (where, O is the origin) and point C is (1, 0, 4), then the value of 102V is equal to

Answer

To solve the problem, we need to find the volume of the tetrahedron formed by the points O (the origin), A, B, and C, where the points A and B are the intersections of the given line with the xy-plane and yz-plane respectively, and C is given as (1, 0, 4).

### Step 1: Find the coordinates of point A (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 point A, we set \( z = 0 \) (since A lies in the xy-plane).

Let \( \lambda \) be the common ratio:
\[ \frac{x-1}{2} = \frac{y-2}{3} = \frac{0-4}{4} \]
From \( \frac{0-4}{4} = -1 \), we have \( \lambda = -1 \).

Now substituting \( \lambda = -1 \) into the equations for x and y:
\[ x = 2(-1) + 1 = -1 \]
\[ y = 3(-1) + 2 = -1 \]
Thus, the coordinates of point A are:
\[ A(-1, -1, 0) \]

### Step 2: Find the coordinates of point B (intersection with the yz-plane)

To find point B, we set \( x = 0 \) (since B lies in the yz-plane).

Let \( \mu \) be the common ratio:
\[ \frac{0-1}{2} = \frac{y-2}{3} = \frac{z-4}{4} \]
From \( \frac{0-1}{2} = -\frac{1}{2} \), we have \( \mu = -\frac{1}{2} \).

Now substituting \( \mu = -\frac{1}{2} \) into the equations for y and z:
\[ y = 3(-\frac{1}{2}) + 2 = 1/2 \]
\[ z = 4(-\frac{1}{2}) + 4 = 2 \]
Thus, the coordinates of point B are:
\[ B(0, \frac{1}{2}, 2) \]

### Step 3: Identify point C

The coordinates of point C are given as:
\[ C(1, 0, 4) \]

### Step 4: Calculate the volume of tetrahedron OABC

The volume \( V \) of the tetrahedron formed by points O, A, B, and C can be calculated using the determinant formula:
\[ V = \frac{1}{6} \left| \begin{vmatrix} x_A & y_A & z_A \\ x_B & y_B & z_B \\ x_C & y_C & z_C \end{vmatrix} \right| \]
Where \( O(0, 0, 0) \), \( A(-1, -1, 0) \), \( B(0, \frac{1}{2}, 2) \), and \( C(1, 0, 4) \).

The matrix becomes:
\[ \begin{vmatrix} -1 & -1 & 0 \\ 0 & \frac{1}{2} & 2 \\ 1 & 0 & 4 \end{vmatrix} \]

Calculating the determinant:
\[ = -1 \left( \frac{1}{2} \cdot 4 - 0 \cdot 2 \right) - (-1) \left( 0 \cdot 4 - 1 \cdot 2 \right) + 0 \left( 0 \cdot 0 - 1 \cdot \frac{1}{2} \right) \]
\[ = -1 \cdot 2 + 1 \cdot 2 = -2 + 2 = 0 \]

Calculating the volume:
\[ V = \frac{1}{6} \cdot 0 = 0 \]

### Step 5: Calculate \( 102V \)

Since \( V = 0 \):
\[ 102V = 102 \cdot 0 = 0 \]

### Final Answer
The value of \( 102V \) is:
\[ \boxed{0} \]