A plane containing the point (3, 2, 0) and the line `(x-1)/(1)=(y-2)/(5)=(z-3)/(4)` also contains the point:

Options

A. `(0,-3,1)`
B. `(0,7,10)`
C. `(0,7,-10)`
D. `(0,3,1)`

Answer

To solve the problem, we need to find the equation of the plane that contains the point (3, 2, 0) and the given line. The line is represented in symmetric form as:

\[ \frac{x-1}{1} = \frac{y-2}{5} = \frac{z-3}{4} \]

### Step 1: Identify the direction ratios of the line
From the line's equation, we can extract the direction ratios. The direction ratios are given by the coefficients of the parameters in the symmetric equation:

- Direction ratios (d1, d2, d3) = (1, 5, 4)

### Step 2: Identify a point on the line
The line passes through the point (1, 2, 3). This point can be obtained by setting the parameter to 0 in the symmetric equation.

### Step 3: Create vectors in the plane
We now have two points in the plane: (3, 2, 0) and (1, 2, 3). We can create a vector from these points:

\[ \text{Vector A} = (3 - 1, 2 - 2, 0 - 3) = (2, 0, -3) \]

We also have the direction vector of the line:

\[ \text{Vector B} = (1, 5, 4) \]

### Step 4: Find the normal vector to the plane
To find the normal vector to the plane, we take the cross product of vectors A and B.

\[ \text{Vector A} = \begin{pmatrix} 2 \\ 0 \\ -3 \end{pmatrix}, \quad \text{Vector B} = \begin{pmatrix} 1 \\ 5 \\ 4 \end{pmatrix} \]

The cross product \( \text{A} \times \text{B} \) is calculated as follows:

\[ \text{A} \times \text{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & -3 \\ 1 & 5 & 4 \end{vmatrix} \]

Calculating the determinant:

\[ = \hat{i}(0 \cdot 4 - (-3) \cdot 5) - \hat{j}(2 \cdot 4 - (-3) \cdot 1) + \hat{k}(2 \cdot 5 - 0 \cdot 1) \]

\[ = \hat{i}(0 + 15) - \hat{j}(8 + 3) + \hat{k}(10) \]

\[ = 15\hat{i} - 11\hat{j} + 10\hat{k} \]

Thus, the normal vector \( \text{N} = (15, -11, 10) \).

### Step 5: Write the equation of the plane
The equation of the plane can be written using the point-normal form:

\[ A(x - x_1) + B(y - y_1) + C(z - z_1) = 0 \]

Substituting \( A = 15, B = -11, C = 10 \) and the point \( (3, 2, 0) \):

\[ 15(x - 3) - 11(y - 2) + 10(z - 0) = 0 \]

Expanding this:

\[ 15x - 45 - 11y + 22 + 10z = 0 \]

Rearranging gives:

\[ 15x - 11y + 10z - 23 = 0 \]

### Step 6: Check the given points
Now we need to check which of the given points satisfy this equation. The options are:

1. \( (0, -3, 1) \)
2. \( (0, 7, 10) \)
3. \( (0, 3, -1) \)
4. \( (0, 3, 1) \)

#### Checking point (0, 7, 10):
Substituting \( (0, 7, 10) \):

\[ 15(0) - 11(7) + 10(10) - 23 = 0 \]
\[ 0 - 77 + 100 - 23 = 0 \]
\[ 0 = 0 \quad \text{(True)} \]

#### Checking point (0, -3, 1):
Substituting \( (0, -3, 1) \):

\[ 15(0) - 11(-3) + 10(1) - 23 = 0 \]
\[ 0 + 33 + 10 - 23 = 0 \]
\[ 20 \neq 0 \quad \text{(False)} \]

#### Checking point (0, 3, -1):
Substituting \( (0, 3, -1) \):

\[ 15(0) - 11(3) + 10(-1) - 23 = 0 \]
\[ 0 - 33 - 10 - 23 = 0 \]
\[ -66 \neq 0 \quad \text{(False)} \]

#### Checking point (0, 3, 1):
Substituting \( (0, 3, 1) \):

\[ 15(0) - 11(3) + 10(1) - 23 = 0 \]
\[ 0 - 33 + 10 - 23 = 0 \]
\[ -46 \neq 0 \quad \text{(False)} \]

### Conclusion
The only point that satisfies the equation of the plane is \( (0, 7, 10) \).