Which of the following points lie on the plane containing the line `(x+1)/(-3)=(y-3)/(2)=(z+2)/(1)` and the point (0,7, -7)?

Options

A. (0,0,0)
B. (1, 2, -3)
C. (2,-2, 0)
D. (1,2,3)

Answer

To determine which points lie on the plane containing the line given by the equation \((x+1)/(-3)=(y-3)/(2)=(z+2)/(1)\) and the point \((0,7,-7)\), we will follow these steps:

### Step 1: Identify the Direction Ratios of the Line
The line can be expressed in parametric form. The direction ratios of the line can be extracted from the equation:
\[ \frac{x+1}{-3} = \frac{y-3}{2} = \frac{z+2}{1} \]
From this, we can see that the direction ratios are \((-3, 2, 1)\).

### Step 2: Identify a Point on the Line
From the parametric equations, we can find a point on the line. Setting the parameter \(t = 0\):
\[ x = -1, \quad y = 3, \quad z = -2 \]
Thus, a point on the line is \((-1, 3, -2)\).

### Step 3: Determine the Normal Vector of the Plane
To find the normal vector of the plane, we need a second vector that lies in the plane. We can use the vector from the point \((-1, 3, -2)\) to the point \((0, 7, -7)\):
\[ \text{Vector } B = (0 - (-1), 7 - 3, -7 - (-2)) = (1, 4, -5) \]
Now we have two vectors in the plane:
1. \(\mathbf{A} = (-3, 2, 1)\)
2. \(\mathbf{B} = (1, 4, -5)\)

We can find the normal vector \(\mathbf{N}\) of the plane by taking the cross product \(\mathbf{A} \times \mathbf{B}\).

### Step 4: Calculate the Cross Product
The cross product is calculated as follows:
\[ \mathbf{N} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -3 & 2 & 1 \\ 1 & 4 & -5 \end{vmatrix} \]
Calculating the determinant:
\[ \mathbf{N} = \mathbf{i} \begin{vmatrix} 2 & 1 \\ 4 & -5 \end{vmatrix} - \mathbf{j} \begin{vmatrix} -3 & 1 \\ 1 & -5 \end{vmatrix} + \mathbf{k} \begin{vmatrix} -3 & 2 \\ 1 & 4 \end{vmatrix} \]
Calculating each determinant:
1. \(\mathbf{i} (2 \cdot -5 - 1 \cdot 4) = \mathbf{i} (-10 - 4) = -14\mathbf{i}\)
2. \(-\mathbf{j} (-3 \cdot -5 - 1 \cdot 1) = -\mathbf{j} (15 - 1) = -14\mathbf{j}\)
3. \(\mathbf{k} (-3 \cdot 4 - 2 \cdot 1) = \mathbf{k} (-12 - 2) = -14\mathbf{k}\)

Thus, the normal vector is:
\[ \mathbf{N} = (-14, 14, -14) \]

### Step 5: Equation of the Plane
The equation of the plane can be given in the form:
\[ \mathbf{N} \cdot (\mathbf{r} - \mathbf{r_0}) = 0 \]
Where \(\mathbf{r_0} = (0, 7, -7)\) and \(\mathbf{r} = (x, y, z)\):
\[ -14(x - 0) + 14(y - 7) - 14(z + 7) = 0 \]
Simplifying this gives:
\[ -14x + 14y - 14z + 14 \cdot 7 - 14 \cdot 7 = 0 \implies -14x + 14y - 14z = 0 \implies x - y + z = 0 \]

### Step 6: Check Which Points Satisfy the Plane Equation
Now, we need to check which of the given points satisfy the equation \(x - y + z = 0\).

1. For point \((0, 0, 0)\):
\[ 0 - 0 + 0 = 0 \quad \text{(satisfies)} \]
2. For point \((1, 2, -3)\):
\[ 1 - 2 - 3 = -4 \quad \text{(does not satisfy)} \]
3. For point \((2, -2, 0)\):
\[ 2 - (-2) + 0 = 4 \quad \text{(does not satisfy)} \]

### Conclusion
The only point that lies on the plane is \((0, 0, 0)\).