To solve the problem, we need to find a plane that contains the line formed by the intersection of the two given planes and is parallel to the line defined by the direction ratios \( x/1 = y/1 = z/1 \). We will also check which of the given points lies on this plane.
### Step 1: Find the direction ratios of the line of intersection of the two planes.
The two planes are given by the equations:
1. \( x - 2y + 3z + 2 = 0 \) (Plane 1)
2. \( 2x + 3y - z + 1 = 0 \) (Plane 2)
To find the direction ratios of the line of intersection, we can take the normal vectors of the planes:
- Normal vector of Plane 1: \( \vec{n_1} = (1, -2, 3) \)
- Normal vector of Plane 2: \( \vec{n_2} = (2, 3, -1) \)
The direction ratios of the line of intersection can be found using the cross product:
\[
\vec{d} = \vec{n_1} \times \vec{n_2}
\]
Calculating the cross product:
\[
\vec{d} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & -2 & 3 \\
2 & 3 & -1
\end{vmatrix}
= \hat{i}((-2)(-1) - (3)(3)) - \hat{j}((1)(-1) - (3)(2)) + \hat{k}((1)(3) - (-2)(2))
\]
\[
= \hat{i}(2 - 9) - \hat{j}(-1 - 6) + \hat{k}(3 + 4)
\]
\[
= -7\hat{i} + 7\hat{j} + 7\hat{k}
\]
Thus, the direction ratios of the line of intersection are \( (-7, 7, 7) \).
### Step 2: Find the equation of the plane containing the line and parallel to \( x/1 = y/1 = z/1 \).
The line \( x/1 = y/1 = z/1 \) has direction ratios \( (1, 1, 1) \). The plane we need to find must have a normal vector that is perpendicular to both the direction ratios of the line of intersection \( (-7, 7, 7) \) and the direction ratios \( (1, 1, 1) \).
To find the normal vector \( \vec{n} \) of the required plane, we can take the cross product:
\[
\vec{n} = \vec{d} \times (1, 1, 1)
\]
Calculating the cross product:
\[
\vec{n} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
-7 & 7 & 7 \\
1 & 1 & 1
\end{vmatrix}
= \hat{i}(7 - 7) - \hat{j}(-7 - 7) + \hat{k}(-7 - 7)
\]
\[
= 0\hat{i} + 14\hat{j} - 14\hat{k}
\]
Thus, the normal vector is \( (0, 14, -14) \) or simplified to \( (0, 1, -1) \).
### Step 3: Form the equation of the plane.
Using the point-normal form of the plane equation:
\[
0(x - x_0) + 1(y - y_0) - 1(z - z_0) = 0
\]
This simplifies to:
\[
y - z = d
\]
where \( d \) is a constant determined by a point on the line of intersection. To find \( d \), we can substitute a point from the intersection of the two planes.
### Step 4: Find a point on the line of intersection.
To find a point, we can solve the two plane equations simultaneously. Setting \( z = 0 \):
1. From Plane 1: \( x - 2y + 2 = 0 \) → \( x = 2y - 2 \)
2. From Plane 2: \( 2(2y - 2) + 3y + 1 = 0 \)
\[
4y - 4 + 3y + 1 = 0 \Rightarrow 7y - 3 = 0 \Rightarrow y = \frac{3}{7}
\]
Substituting \( y \) back to find \( x \):
\[
x = 2\left(\frac{3}{7}\right) - 2 = \frac{6}{7} - 2 = \frac{6}{7} - \frac{14}{7} = -\frac{8}{7}
\]
Thus, a point on the line is \( \left(-\frac{8}{7}, \frac{3}{7}, 0\right) \).
### Step 5: Substitute the point into the plane equation.
Now we can find \( d \):
\[
\frac{3}{7} - 0 = d \Rightarrow d = \frac{3}{7}
\]
Thus, the equation of the plane is:
\[
y - z = \frac{3}{7}
\]
### Step 6: Check which points lie on the plane.
Now we will check the given points to see which one satisfies the plane equation \( y - z = \frac{3}{7} \).
1. **Point \( (1, 0, 3/7) \)**:
\[
0 - \frac{3}{7} = -\frac{3}{7} \quad \text{(not on the plane)}
\]
2. **Point \( (0, 5/7, 2/7) \)**:
\[
\frac{5}{7} - \frac{2}{7} = \frac{3}{7} \quad \text{(on the plane)}
\]
3. **Point \( (0, 3/7, -3/7) \)**:
\[
\frac{3}{7} - (-\frac{3}{7}) = \frac{6}{7} \quad \text{(not on the plane)}
\]
4. **Point \( (3, 5/7, -2/7) \)**:
\[
\frac{5}{7} - (-\frac{2}{7}) = \frac{7}{7} = 1 \quad \text{(not on the plane)}
\]
### Conclusion:
The only point that lies on the plane is \( (0, \frac{5}{7}, \frac{2}{7}) \).
