Consider the plane through `(2, 3, -1)` and at right angles to the vector `3hat(i)-4hat(j)+7hat(k)` from the origin is

Options

A. The equation of the plane through the given point is `3x-4y+7z+13=0`.
B. perpendicular distance of plane from origin `(1)/(sqrt(74))`.
C. perpendicular distance of plane from origin `(13)/(sqrt(74))`.
D. perpendicular distance of plane from origin `(21)/(sqrt(74))`.

Answer

To solve the problem, we need to find the equation of a plane that passes through the point \((2, 3, -1)\) and is perpendicular to the vector \(\vec{n} = 3\hat{i} - 4\hat{j} + 7\hat{k}\).

### Step-by-Step Solution:

1. **Identify the point and the normal vector:**
- The point through which the plane passes is given as \(P(2, 3, -1)\).
- The normal vector to the plane is given as \(\vec{n} = 3\hat{i} - 4\hat{j} + 7\hat{k}\), which can be represented as the vector \((3, -4, 7)\).

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 \((x_1, y_1, z_1)\) is a point on the plane and \((a, b, c)\) are the components of the normal vector.

3. **Substituting the values:**
Here, \((x_1, y_1, z_1) = (2, 3, -1)\) and \((a, b, c) = (3, -4, 7)\). Substituting these values into the equation:
\[ 3(x - 2) - 4(y - 3) + 7(z + 1) = 0 \]

4. **Expanding the equation:**
Expanding this equation gives:
\[ 3x - 6 - 4y + 12 + 7z + 7 = 0 \]
Simplifying this:
\[ 3x - 4y + 7z + 13 = 0 \]

5. **Final equation of the plane:**
The equation of the plane is:
\[ 3x - 4y + 7z + 13 = 0 \]

6. **Finding the perpendicular distance from the origin:**
The formula for the perpendicular distance \(D\) from a point \((x_1, y_1, z_1)\) to the plane \(Ax + By + Cz + D = 0\) is:
\[ D = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
Here, the origin is \((0, 0, 0)\), and substituting \(A = 3\), \(B = -4\), \(C = 7\), and \(D = 13\):
\[ D = \frac{|3(0) - 4(0) + 7(0) + 13|}{\sqrt{3^2 + (-4)^2 + 7^2}} = \frac{|13|}{\sqrt{9 + 16 + 49}} = \frac{13}{\sqrt{74}} \]

### Final Results:
- The equation of the plane is:
\[ 3x - 4y + 7z + 13 = 0 \]
- The perpendicular distance from the origin to the plane is:
\[ D = \frac{13}{\sqrt{74}} \]