Find the equation of a plane passing through the intersection of the planes `x-3y+2z-5 - 0` and `2x-y+3z-1 = 0` and passes through the point `(1,-2,3)`.

Answer

To find the equation of a plane passing through the intersection of the planes \( x - 3y + 2z - 5 = 0 \) and \( 2x - y + 3z - 1 = 0 \) and also passing through the point \( (1, -2, 3) \), we can follow these steps:

### Step 1: Write the equation of the plane through the intersection of the two given planes.
The equation of a plane through the intersection of two planes can be expressed as:
\[ \pi: (x - 3y + 2z - 5) + \lambda (2x - y + 3z - 1) = 0 \]
where \( \lambda \) is a parameter.

### Step 2: Substitute the point \( (1, -2, 3) \) into the equation.
We substitute \( x = 1 \), \( y = -2 \), and \( z = 3 \) into the equation:
\[ (1 - 3(-2) + 2(3) - 5) + \lambda (2(1) - (-2) + 3(3) - 1) = 0 \]

### Step 3: Simplify the equation.
Calculating the first part:
\[ 1 + 6 + 6 - 5 = 8 \]
Calculating the second part:
\[ 2 + 2 + 9 - 1 = 12 \]
Thus, we have:
\[ 8 + 12\lambda = 0 \]

### Step 4: Solve for \( \lambda \).
Rearranging gives:
\[ 12\lambda = -8 \implies \lambda = -\frac{8}{12} = -\frac{2}{3} \]

### Step 5: Substitute \( \lambda \) back into the plane equation.
Now substitute \( \lambda = -\frac{2}{3} \) back into the equation of the plane:
\[ (x - 3y + 2z - 5) - \frac{2}{3}(2x - y + 3z - 1) = 0 \]

### Step 6: Simplify the equation.
Expanding the equation:
\[ x - 3y + 2z - 5 - \frac{4}{3}x + \frac{2}{3}y - 2z + \frac{2}{3} = 0 \]
Combining like terms:
\[ \left(1 - \frac{4}{3}\right)x + \left(-3 + \frac{2}{3}\right)y + \left(2 - 2\right)z + \left(-5 + \frac{2}{3}\right) = 0 \]
This simplifies to:
\[ -\frac{1}{3}x - \frac{7}{3}y - \frac{13}{3} = 0 \]

### Step 7: Multiply through by -3 to eliminate fractions.
Multiplying through by -3 gives:
\[ x + 7y + 13 = 0 \]

### Final Answer:
The equation of the required plane is:
\[ \boxed{x + 7y + 13 = 0} \]