A plane `2x+3y+5z=1` has a point P which is at minimum distance from line joining `A(1, 0, -3), B(1, -5, 7),` then distance AP is equal to

Options

A. `3sqrt(5)`
B. `2sqrt(5)`
C. `4sqrt(4)`
D. None of these

Answer

To solve the problem, we need to find the distance from point A(1, 0, -3) to the point P on the plane \(2x + 3y + 5z = 1\) that is at a minimum distance from the line joining points A and B(1, -5, 7).

### Step-by-Step Solution:

1. **Find the Direction Vector of Line AB**:
The direction vector of the line joining points A and B can be calculated as:
\[ \vec{AB} = B - A = (1 - 1, -5 - 0, 7 - (-3)) = (0, -5, 10) \]

2. **Parametric Equations of Line AB**:
The parametric equations of the line can be expressed as:
\[ x = 1, \quad y = -5t, \quad z = 10t - 3 \]
where \(t\) is a parameter.

3. **Point Q on Line AB**:
From the parametric equations, we can express the coordinates of point Q as:
\[ Q(1, -5t, 10t - 3) \]

4. **Substituting Q into the Plane Equation**:
To find the minimum distance, we need to substitute the coordinates of Q into the plane equation \(2x + 3y + 5z = 1\):
\[ 2(1) + 3(-5t) + 5(10t - 3) = 1 \]
Simplifying this gives:
\[ 2 - 15t + 50t - 15 = 1 \implies 35t - 13 = 1 \implies 35t = 14 \implies t = \frac{2}{5} \]

5. **Finding Coordinates of Point Q**:
Substitute \(t = \frac{2}{5}\) back into the parametric equations to find the coordinates of Q:
\[ Q\left(1, -5\left(\frac{2}{5}\right), 10\left(\frac{2}{5}\right) - 3\right) = \left(1, -2, 1\right) \]

6. **Point P and Q Coincide**:
Since point P lies on the plane and is at minimum distance from line AB, we find that \(P = Q = (1, -2, 1)\).

7. **Calculating Distance AP**:
Now, we calculate the distance from point A(1, 0, -3) to point P(1, -2, 1) using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
Substituting the coordinates:
\[ d = \sqrt{(1 - 1)^2 + (-2 - 0)^2 + (1 - (-3))^2} = \sqrt{0 + 4 + 16} = \sqrt{20} = 2\sqrt{5} \]

### Final Answer:
The distance \(AP\) is equal to \(2\sqrt{5}\) units.