The distance of the point `(-1, -5, -10)` from the point of intersection of the line `(x-2)/(2)=(y+1)/(4)=(z-2)/(12)` and the plane `x-y+z=5` is

Options

A. `2sqrt(11)`
B. `sqrt(126)`
C. `13`
D. `14`

Answer

To solve the problem, we need to find the distance from the point \((-1, -5, -10)\) to the point of intersection of the given line and the plane.

### Step 1: Parametrize the Line
The line is given in the symmetric form:
\[ \frac{x - 2}{2} = \frac{y + 1}{4} = \frac{z - 2}{12} \]
Let \( t = \frac{x - 2}{2} = \frac{y + 1}{4} = \frac{z - 2}{12} \). We can express \(x\), \(y\), and \(z\) in terms of \(t\):
\[ x = 2t + 2, \quad y = 4t - 1, \quad z = 12t + 2 \]

### Step 2: Substitute into the Plane Equation
The equation of the plane is:
\[ x - y + z = 5 \]
Substituting the parametric equations of the line into the plane equation:
\[ (2t + 2) - (4t - 1) + (12t + 2) = 5 \]
Simplifying this:
\[ 2t + 2 - 4t + 1 + 12t + 2 = 5 \]
\[ (2t - 4t + 12t) + (2 + 1 + 2) = 5 \]
\[ 10t + 5 = 5 \]
\[ 10t = 0 \implies t = 0 \]

### Step 3: Find the Point of Intersection
Now, substitute \(t = 0\) back into the parametric equations to find the coordinates of the intersection point:
\[ x = 2(0) + 2 = 2, \quad y = 4(0) - 1 = -1, \quad z = 12(0) + 2 = 2 \]
Thus, the point of intersection is \((2, -1, 2)\).

### Step 4: Calculate the Distance
Now, we need to find the distance from the point \((-1, -5, -10)\) to the point \((2, -1, 2)\). The distance formula in 3D is given by:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
Substituting the points:
\[ d = \sqrt{(2 - (-1))^2 + (-1 - (-5))^2 + (2 - (-10))^2} \]
Calculating each term:
\[ = \sqrt{(2 + 1)^2 + (-1 + 5)^2 + (2 + 10)^2} \]
\[ = \sqrt{(3)^2 + (4)^2 + (12)^2} \]
\[ = \sqrt{9 + 16 + 144} \]
\[ = \sqrt{169} = 13 \]

### Final Answer
The distance from the point \((-1, -5, -10)\) to the point of intersection is \(13\) units.

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