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 given by the equations \( \pi_1: x - 3y + 2z - 5 = 0 \) and \( \pi_2: 2x + y + 3z - 1 = 0 \), and also passing through the point \( (1, -2, 3) \), we can follow these steps:

### Step 1: Set up the equation of the plane
The equation of a plane that passes through the intersection of two planes can be expressed as:
\[ \pi_1 + \lambda \pi_2 = 0 \]
where \( \lambda \) is a parameter. Substituting the equations of the planes, we have:
\[ (x - 3y + 2z - 5) + \lambda(2x + y + 3z - 1) = 0 \]

### Step 2: Expand the equation
Expanding the equation gives:
\[ x - 3y + 2z - 5 + \lambda(2x + y + 3z - 1) = 0 \]
This can be rewritten as:
\[ x - 3y + 2z - 5 + 2\lambda x + \lambda y + 3\lambda z - \lambda = 0 \]
Combining like terms, we get:
\[ (1 + 2\lambda)x + (-3 + \lambda)y + (2 + 3\lambda)z - (5 + \lambda) = 0 \]

### Step 3: Substitute the point into the equation
Now we need to substitute the point \( (1, -2, 3) \) into the equation to find the value of \( \lambda \):
\[ (1 + 2\lambda)(1) + (-3 + \lambda)(-2) + (2 + 3\lambda)(3) - (5 + \lambda) = 0 \]
Calculating this gives:
\[ (1 + 2\lambda) + (6 - 2\lambda) + (6 + 9\lambda) - (5 + \lambda) = 0 \]
Simplifying this:
\[ 1 + 2\lambda + 6 - 2\lambda + 6 + 9\lambda - 5 - \lambda = 0 \]
Combining all terms:
\[ (1 + 6 + 6 - 5) + (2\lambda - 2\lambda + 9\lambda - \lambda) = 0 \]
This simplifies to:
\[ 8 + 8\lambda = 0 \]

### Step 4: Solve for \( \lambda \)
From the equation \( 8 + 8\lambda = 0 \), we can solve for \( \lambda \):
\[ 8\lambda = -8 \implies \lambda = -1 \]

### Step 5: Substitute \( \lambda \) back into the plane equation
Now we substitute \( \lambda = -1 \) back into the equation of the plane:
\[ (1 + 2(-1))x + (-3 + (-1))y + (2 + 3(-1))z - (5 + (-1)) = 0 \]
This simplifies to:
\[ (1 - 2)x + (-3 - 1)y + (2 - 3)z - (5 - 1) = 0 \]
Which gives:
\[ -x - 4y - z - 4 = 0 \]

### Step 6: Rearranging the equation
Rearranging the above equation gives us the final equation of the plane:
\[ x + 4y + z + 4 = 0 \]

### Final Answer
The equation of the plane is:
\[ \boxed{x + 4y + z + 4 = 0} \]