The equation of the plane which passes through the point of intersection of `(x-1)/(3)=(y-2)/(1)=(z-3)/(2) and (x-3)/(1)=(y-1)/(2)=(z-2)/(3)` and perpendicular to `4hati+3hatj+5hatk`, is

Options

A. `4x+3y+5z=50`
B. `4x+3y+5x=25`
C. `4x+3y+5z=10`
D. `4x-3y+5z=50`

Answer

To find the equation of the plane that passes through the point of intersection of the given lines and is perpendicular to the vector \(4\hat{i} + 3\hat{j} + 5\hat{k}\), we will follow these steps:

### Step 1: Find the equations of the lines
The first line is given by:
\[ \frac{x-1}{3} = \frac{y-2}{1} = \frac{z-3}{2} \]
Let \( \lambda \) be the parameter for this line. We can express the coordinates \( (x, y, z) \) in terms of \( \lambda \):
\[ x = 3\lambda + 1, \quad y = \lambda + 2, \quad z = 2\lambda + 3 \]
So, the parametric equations for line \( L_1 \) are:
\[ (x, y, z) = (3\lambda + 1, \lambda + 2, 2\lambda + 3) \]

The second line is given by:
\[ \frac{x-3}{1} = \frac{y-1}{2} = \frac{z-2}{3} \]
Let \( \mu \) be the parameter for this line. The coordinates can be expressed as:
\[ x = \mu + 3, \quad y = 2\mu + 1, \quad z = 3\mu + 2 \]
So, the parametric equations for line \( L_2 \) are:
\[ (x, y, z) = (\mu + 3, 2\mu + 1, 3\mu + 2) \]

### Step 2: Set the coordinates equal to find the intersection point
At the point of intersection \( P \), the coordinates from both lines must be equal:
\[ 3\lambda + 1 = \mu + 3 \quad (1) \]
\[ \lambda + 2 = 2\mu + 1 \quad (2) \]
\[ 2\lambda + 3 = 3\mu + 2 \quad (3) \]

### Step 3: Solve the equations
From equation (1):
\[ 3\lambda - \mu = 2 \quad (4) \]
From equation (2):
\[ \lambda - 2\mu = -1 \quad (5) \]

Now, we can solve equations (4) and (5). From (4), we can express \( \mu \):
\[ \mu = 3\lambda - 2 \]
Substituting \( \mu \) into (5):
\[ \lambda - 2(3\lambda - 2) = -1 \]
\[ \lambda - 6\lambda + 4 = -1 \]
\[ -5\lambda + 4 = -1 \]
\[ -5\lambda = -5 \implies \lambda = 1 \]

Now substituting \( \lambda = 1 \) back into (4) to find \( \mu \):
\[ \mu = 3(1) - 2 = 1 \]

### Step 4: Find the coordinates of the intersection point \( P \)
Substituting \( \lambda = 1 \) into the parametric equations of \( L_1 \):
\[ x = 3(1) + 1 = 4, \quad y = 1 + 2 = 3, \quad z = 2(1) + 3 = 5 \]
Thus, the coordinates of point \( P \) are \( (4, 3, 5) \).

### Step 5: Write the equation of the plane
The normal vector of the plane is given as \( \mathbf{n} = 4\hat{i} + 3\hat{j} + 5\hat{k} \). The equation of the plane can be written in the form:
\[ \mathbf{n} \cdot \mathbf{r} = d \]
Where \( \mathbf{r} = x\hat{i} + y\hat{j} + z\hat{k} \) and \( d \) is determined using point \( P(4, 3, 5) \):
\[ 4(x - 4) + 3(y - 3) + 5(z - 5) = 0 \]
Expanding this:
\[ 4x - 16 + 3y - 9 + 5z - 25 = 0 \]
\[ 4x + 3y + 5z - 50 = 0 \]
Thus, the equation of the plane is:
\[ 4x + 3y + 5z = 50 \]

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