Find the equation of the plane passing through the point(0, 7,-7) and containing the line `(x+1)/(-3)=(y-3)/2=(z+2)/1`

Answer

To find the equation of the plane passing through the point \( (0, 7, -7) \) and containing the line given by the equations \( \frac{x+1}{-3} = \frac{y-3}{2} = \frac{z+2}{1} \), we can follow these steps:

### Step 1: Identify a point on the line
The line can be expressed in parametric form. Let \( t \) be the parameter:
\[ x = -3t - 1, \quad y = 2t + 3, \quad z = t - 2 \]
When \( t = 0 \), we find a point on the line:
\[ x = -1, \quad y = 3, \quad z = -2 \]
So, the point \( (-1, 3, -2) \) lies on the line.

### Step 2: Find the direction vector of the line
From the parametric equations, we can determine the direction vector of the line:
\[ \vec{d} = (-3, 2, 1) \]

### Step 3: Find the normal vector of the plane
The plane must contain the point \( (0, 7, -7) \) and the point \( (-1, 3, -2) \). We can find a vector in the plane by subtracting these two points:
\[ \vec{v} = (-1 - 0, 3 - 7, -2 + 7) = (-1, -4, 5) \]

### Step 4: Find the normal vector using cross product
The normal vector \( \vec{n} \) of the plane can be found by taking the cross product of the direction vector \( \vec{d} \) and the vector \( \vec{v} \):
\[ \vec{n} = \vec{d} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 2 & 1 \\ -1 & -4 & 5 \end{vmatrix} \]
Calculating the determinant:
\[ \vec{n} = \hat{i} \begin{vmatrix} 2 & 1 \\ -4 & 5 \end{vmatrix} - \hat{j} \begin{vmatrix} -3 & 1 \\ -1 & 5 \end{vmatrix} + \hat{k} \begin{vmatrix} -3 & 2 \\ -1 & -4 \end{vmatrix} \]
Calculating each of the 2x2 determinants:
1. \( \begin{vmatrix} 2 & 1 \\ -4 & 5 \end{vmatrix} = (2)(5) - (1)(-4) = 10 + 4 = 14 \)
2. \( \begin{vmatrix} -3 & 1 \\ -1 & 5 \end{vmatrix} = (-3)(5) - (1)(-1) = -15 + 1 = -14 \)
3. \( \begin{vmatrix} -3 & 2 \\ -1 & -4 \end{vmatrix} = (-3)(-4) - (2)(-1) = 12 + 2 = 14 \)

Thus,
\[ \vec{n} = (14)\hat{i} + (14)\hat{j} + (14)\hat{k} = (14, 14, 14) \]

### Step 5: Write the equation of the plane
The equation of the plane can be expressed as:
\[ 14(x - 0) + 14(y - 7) + 14(z + 7) = 0 \]
This simplifies to:
\[ 14x + 14y - 98 + 14z + 98 = 0 \]
Thus, we have:
\[ 14x + 14y + 14z = 0 \]
Dividing through by 14 gives:
\[ x + y + z = 0 \]

### Final Answer
The equation of the plane is:
\[ x + y + z = 0 \]