The equation of the plane which passes through (2,-3,1) and is normal to the line joining the points (3,4,-1) and (2,-1,5) is given by

Options

A. `x+5y-6z+19=0`
B. `x-5y+6z-19=0`
C. `x+5y+6z+19=0`
D. `x-5y-6z-19=0`

Explanation

To find the equation of the plane that passes through the point (2, -3, 1) and is normal to the line joining the points (3, 4, -1) and (2, -1, 5), we can follow these steps:

### Step 1: Find the direction ratios of the line
The direction ratios of the line joining the points (3, 4, -1) and (2, -1, 5) can be calculated as follows:

- Direction ratio in x = 2 - 3 = -1
- Direction ratio in y = -1 - 4 = -5
- Direction ratio in z = 5 - (-1) = 6

Thus, the direction ratios (a, b, c) are (-1, -5, 6).

### Step 2: Use the point-normal form of the plane equation
The equation of a plane in point-normal form is given by:
\[ a(x - x_1) + b(y - y_1) + c(z - z_1) = 0 \]
where (x1, y1, z1) is a point on the plane and (a, b, c) are the direction ratios of the normal to the plane.

Here, we have:
- Point (x1, y1, z1) = (2, -3, 1)
- Direction ratios (a, b, c) = (-1, -5, 6)

### Step 3: Substitute the values into the equation
Substituting the values into the equation:
\[ -1(x - 2) - 5(y + 3) + 6(z - 1) = 0 \]

### Step 4: Simplify the equation
Now, let's simplify the equation:
\[ -1(x - 2) - 5(y + 3) + 6(z - 1) = 0 \]
Expanding this:
\[ -x + 2 - 5y - 15 + 6z - 6 = 0 \]
Combining like terms:
\[ -x - 5y + 6z - 19 = 0 \]
Rearranging gives us:
\[ x + 5y - 6z + 19 = 0 \]

### Final Equation of the Plane
Thus, the equation of the plane is:
\[ x + 5y - 6z + 19 = 0 \]

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