Find the equation of the plane passing through the line of intersection of the planes `x + 2y + 3z- 5= 0` and `3x - 2y - z + 1 = 0` and cutting off equal intercepts on the X and Z-axes.

Answer

To find the equation of the plane passing through the line of intersection of the planes \( x + 2y + 3z - 5 = 0 \) and \( 3x - 2y - z + 1 = 0 \) and cutting off equal intercepts on the X and Z-axes, we can follow these steps:

### Step 1: Write the equation of the plane
The equation of a plane passing through the line of intersection of two planes can be expressed as:
\[ P_1 + kP_2 = 0 \]
where \( P_1 \) and \( P_2 \) are the equations of the given planes, and \( k \) is a constant.

Here, we have:
- \( P_1: x + 2y + 3z - 5 = 0 \)
- \( P_2: 3x - 2y - z + 1 = 0 \)

Thus, the equation of the required plane is:
\[ (x + 2y + 3z - 5) + k(3x - 2y - z + 1) = 0 \]

### Step 2: Simplify the equation
Expanding the equation gives:
\[ x + 2y + 3z - 5 + k(3x - 2y - z + 1) = 0 \]
\[ x + 2y + 3z - 5 + 3kx - 2ky - kz + k = 0 \]
Combining like terms:
\[ (1 + 3k)x + (2 - 2k)y + (3 - k)z + (k - 5) = 0 \]

### Step 3: Find the intercepts
To find the x-intercept, set \( y = 0 \) and \( z = 0 \):
\[ (1 + 3k)x + (k - 5) = 0 \]
Thus, the x-intercept \( a \) is:
\[ x = \frac{5 - k}{1 + 3k} \]

To find the z-intercept, set \( x = 0 \) and \( y = 0 \):
\[ (3 - k)z + (k - 5) = 0 \]
Thus, the z-intercept \( c \) is:
\[ z = \frac{5 - k}{3 - k} \]

### Step 4: Set the intercepts equal
Since the problem states that the x-intercept and z-intercept are equal, we set:
\[ \frac{5 - k}{1 + 3k} = \frac{5 - k}{3 - k} \]

Assuming \( 5 - k \neq 0 \) (we will check this later), we can cancel \( 5 - k \) from both sides:
\[ \frac{1}{1 + 3k} = \frac{1}{3 - k} \]

Cross-multiplying gives:
\[ 3 - k = 1 + 3k \]
\[ 3 - 1 = 4k \]
\[ 2 = 4k \implies k = \frac{1}{2} \]

### Step 5: Substitute \( k \) back into the plane equation
Substituting \( k = \frac{1}{2} \) back into the equation of the plane:
\[ (1 + 3 \cdot \frac{1}{2})x + (2 - 2 \cdot \frac{1}{2})y + (3 - \frac{1}{2})z + (\frac{1}{2} - 5) = 0 \]
This simplifies to:
\[ \frac{5}{2}x + y + \frac{5}{2}z - \frac{9}{2} = 0 \]
Multiplying through by 2 to eliminate the fractions:
\[ 5x + 2y + 5z - 9 = 0 \]

### Final Answer
The equation of the required plane is:
\[ 5x + 2y + 5z - 9 = 0 \]