A straight line passes through `(1,-2,3)` and perpendicular to the plane `2x+3y-z=7`
What are the direction rations of normal to plane ?

Options

A. `lt2,3,-1gt`
B. `lt2,3,1gt`
C. `lt-1,2,3gt`
D. None of these

Answer

To find the direction ratios of the normal to the plane given by the equation \(2x + 3y - z = 7\), we can follow these steps:

### Step 1: Identify the coefficients of the plane equation
The general equation of a plane can be expressed as:
\[ ax + by + cz + d = 0 \]
In our case, the given equation is:
\[ 2x + 3y - z - 7 = 0 \]
From this equation, we can identify the coefficients:
- \(a = 2\)
- \(b = 3\)
- \(c = -1\)

### Step 2: Determine the direction ratios of the normal
The direction ratios of the normal to the plane are simply the coefficients \(a\), \(b\), and \(c\). Therefore, the direction ratios of the normal to the plane are:
\[ (2, 3, -1) \]

### Conclusion
Thus, the direction ratios of the normal to the plane \(2x + 3y - z = 7\) are \(2, 3, -1\).

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