The equation of the plane which passes through the point of intersection of lines `""(x-1)/3=(y-2)/1=(z-3)/2"","and"(x-3)/1=(y-1)/2=(z-2)/3` and at greatest distance from point `(0,0,0)` is a. `4x+3y+5z=25` b. `4x+3y=5z=50` c. `3x+4y+5z=49` d. `x+7y-5z=2`

Options

A. `4x+3y+5z=25`
B. `4x+3y+5z=50`
C. `3x+4y+5z=49`
D. `x+7y+5z=2`

Explanation

To find the equation of the plane that passes through the point of intersection of the given lines and is at the greatest distance from the origin, we can follow these steps:

### Step 1: Find the parametric equations of the lines

The first line is given by:
\[ \frac{x-1}{3} = \frac{y-2}{1} = \frac{z-3}{2} \]
Let \( t = \frac{x-1}{3} = \frac{y-2}{1} = \frac{z-3}{2} \). Then we can express the coordinates as:
\[ x = 3t + 1, \quad y = t + 2, \quad z = 2t + 3 \]

The second line is given by:
\[ \frac{x-3}{1} = \frac{y-1}{2} = \frac{z-2}{3} \]
Let \( s = \frac{x-3}{1} = \frac{y-1}{2} = \frac{z-2}{3} \). Then we can express the coordinates as:
\[ x = s + 3, \quad y = 2s + 1, \quad z = 3s + 2 \]

### Step 2: Set the equations equal to find the point of intersection

We equate the coordinates from both lines:
1. \( 3t + 1 = s + 3 \)
2. \( t + 2 = 2s + 1 \)
3. \( 2t + 3 = 3s + 2 \)

### Step 3: Solve the equations

From the first equation:
\[ 3t - s = 2 \quad \text{(Equation 1)} \]
From the second equation:
\[ t - 2s = -1 \quad \text{(Equation 2)} \]
From the third equation:
\[ 2t - 3s = -1 \quad \text{(Equation 3)} \]

Now we can solve these equations.

From Equation 1, we can express \( s \):
\[ s = 3t - 2 \]

Substituting \( s \) in Equation 2:
\[ t - 2(3t - 2) = -1 \implies t - 6t + 4 = -1 \implies -5t + 4 = -1 \implies -5t = -5 \implies t = 1 \]

Now substituting \( t = 1 \) back to find \( s \):
\[ s = 3(1) - 2 = 1 \]

### Step 4: Find the coordinates of the point of intersection

Substituting \( t = 1 \) into the first line's parametric equations:
\[ x = 3(1) + 1 = 4, \quad y = 1 + 2 = 3, \quad z = 2(1) + 3 = 5 \]

Thus, the point of intersection is \( (4, 3, 5) \).

### Step 5: Find the normal vector for the plane at greatest distance from the origin

To maximize the distance from the origin, the plane must be perpendicular to the line connecting the origin to the point \( (4, 3, 5) \). The normal vector \( \vec{n} \) of the plane can be taken as the vector from the origin to this point:
\[ \vec{n} = (4, 3, 5) \]

### Step 6: Write the equation of the plane

The general equation of a plane is given by:
\[ Ax + By + Cz = D \]
Substituting the normal vector components:
\[ 4x + 3y + 5z = D \]

To find \( D \), we substitute the point \( (4, 3, 5) \):
\[ 4(4) + 3(3) + 5(5) = D \implies 16 + 9 + 25 = D \implies D = 50 \]

Thus, the equation of the plane is:
\[ 4x + 3y + 5z = 50 \]

### Final Answer
The equation of the plane is:
\[ \boxed{4x + 3y + 5z = 50} \]