The equation of the plane passing through the point (0, 7, -7) and containing the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1},$ is:

Question

Accepted Answer

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

1. **Identify the Direction Vector of the Line:**
   The line is given in symmetric form: $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}=t$. From this, we can extract the direction vector of the line as $\mathbf{d} = \langle -3, 2, 1 \rangle$.

2. **Find a Point on the Line:**
   By setting $t = 0$, we find a point on the line: 
   $$
   x = -1, \quad y = 3, \quad z = -2
   $$
   So, a point on the line is $(-1, 3, -2)$.

3. **Determine the Normal Vector of the Plane:**
   The plane must contain both the point $(0, 7, -7)$ and the line. Therefore, the normal vector of the plane, $\mathbf{n}$, must be perpendicular to both the direction vector of the line $\mathbf{d} = \langle -3, 2, 1 \rangle$ and the vector connecting the point $(0, 7, -7)$ to the point $(-1, 3, -2)$ on the line. This vector is:
   $$
   \mathbf{v} = \langle -1 - 0, 3 - 7, -2 + 7 \rangle = \langle -1, -4, 5 \rangle
   $$

4. **Calculate the Cross Product:**
   The normal vector $\mathbf{n}$ can be found by taking the cross product of $\mathbf{d}$ and $\mathbf{v}$:
   $$
   \mathbf{n} = \mathbf{d} \times \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -3 & 2 & 1 \ -1 & -4 & 5 \end{vmatrix}
   $$
   $$
   = \mathbf{i}(2 \cdot 5 - 1 \cdot (-4)) - \mathbf{j}(-3 \cdot 5 - 1 \cdot (-1)) + \mathbf{k}(-3 \cdot (-4) - 2 \cdot (-1))
   $$
   $$
   = \mathbf{i}(10 + 4) - \mathbf{j}(-15 + 1) + \mathbf{k}(12 + 2)
   $$
   $$
   = \mathbf{i}(14) - \mathbf{j}(-14) + \mathbf{k}(14)
   $$
   $$
   = \langle 14, 14, 14 \rangle
   $$

5. **Equation of the Plane:**
   The equation of the plane is given by:
   $$
   14(x - 0) + 14(y - 7) + 14(z + 7) = 0
   $$
   Simplifying, we get:
   $$
   14x + 14y - 98 + 14z + 98 = 0
   $$
   $$
   14x + 14y + 14z = 0
   $$
   Dividing through by 14:
   $$
   x + y + z = 0
   $$

Thus, the equation of the plane is $x + y + z = 0$, which corresponds to Option 3.

**Explanation for Incorrect Options:**

- **Option 1: $x + 7y + 7z = 0$**
  - This equation does not satisfy the condition of passing through the point (0, 7, -7) and containing the line. Substituting the point (0, 7, -7) into this equation does not yield a true statement.

- **Option 2: $x + 2y + z = 7$**
  - Similarly, substituting the point (0, 7, -7) into this equation does not satisfy it, and the direction vector of the line does not lie on this plane.

- **Option 4: $x + y - z = 14$**
  - This equation also fails to satisfy the condition of passing through the point (0, 7, -7) and containing the line. Substituting the point into this equation does not yield a true statement.

Therefore, Option 3 is the correct answer as it satisfies all the conditions required for the plane.