Find the direction ratios of the normal to the plane `x+2y-3z=5`.

Explanation

To find the direction ratios of the normal to the plane given by the equation \( x + 2y - 3z = 5 \), we can follow these steps:

### Step 1: Identify the general form of the plane equation
The general equation of a plane in three-dimensional space can be expressed as:
\[ Ax + By + Cz = D \]
where \( A \), \( B \), and \( C \) are the coefficients that represent the direction ratios of the normal to the plane.

### Step 2: Compare the given equation with the general form
The given equation of the plane is:
\[ x + 2y - 3z = 5 \]
We can rewrite this in the form \( Ax + By + Cz = D \) by identifying the coefficients of \( x \), \( y \), and \( z \).

### Step 3: Extract coefficients
From the equation \( x + 2y - 3z = 5 \), we can see that:
- The coefficient of \( x \) (A) is \( 1 \)
- The coefficient of \( y \) (B) is \( 2 \)
- The coefficient of \( z \) (C) is \( -3 \)

### Step 4: Write the direction ratios
The direction ratios of the normal to the plane are given by the coefficients \( A \), \( B \), and \( C \). Therefore, the direction ratios of the normal to the plane \( x + 2y - 3z = 5 \) are:
\[ 1, 2, -3 \]

### Final Answer
The direction ratios of the normal to the plane \( x + 2y - 3z = 5 \) are \( (1, 2, -3) \).

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