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

Options

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

Explanation

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-by-Step Solution:

1. **Identify the equation of the plane**: The equation of the plane is given as \(2x + 3y - z = 7\).

2. **Rewrite the plane equation in standard form**: We can rewrite the equation in the form \(Ax + By + Cz + D = 0\). This gives us:
\[ 2x + 3y - z - 7 = 0 \]
Here, \(A = 2\), \(B = 3\), \(C = -1\), and \(D = -7\).

3. **Extract the direction ratios**: The direction ratios of the normal to the plane can be directly obtained from the coefficients of \(x\), \(y\), and \(z\) in the plane equation. Thus, the direction ratios are:
\[ (A, B, C) = (2, 3, -1) \]

4. **Conclusion**: Therefore, the direction ratios of the normal to the plane are \(2, 3, -1\).

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

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