Find the distance of the points `(-1, -5, -10)` form the point of intersection of the line `(x-2)/(3)=(y+1)/(4)=(z-2)/(12)` and plane `x-y+z=5`

Answer

To find the distance of the point \((-1, -5, -10)\) from the point of intersection of the line given by the equations \(\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12}\) and the plane \(x - y + z = 5\), we will follow these steps:

### Step 1: Parametrize the Line
The line can be expressed in parametric form. Let \(\lambda\) be the parameter:
- \(x = 3\lambda + 2\)
- \(y = 4\lambda - 1\)
- \(z = 12\lambda + 2\)

### Step 2: Substitute into the Plane Equation
Now we substitute these parametric equations into the plane equation \(x - y + z = 5\):
\[ (3\lambda + 2) - (4\lambda - 1) + (12\lambda + 2) = 5 \]
Simplifying this:
\[ 3\lambda + 2 - 4\lambda + 1 + 12\lambda + 2 = 5 \]
Combine like terms:
\[ (3\lambda - 4\lambda + 12\lambda) + (2 + 1 + 2) = 5 \]
\[ 11\lambda + 5 = 5 \]

### Step 3: Solve for \(\lambda\)
Now, we solve for \(\lambda\):
\[ 11\lambda = 5 - 5 \]
\[ 11\lambda = 0 \implies \lambda = 0 \]

### Step 4: Find the Intersection Point
Substituting \(\lambda = 0\) back into the parametric equations to find the coordinates of the intersection point:
- \(x = 3(0) + 2 = 2\)
- \(y = 4(0) - 1 = -1\)
- \(z = 12(0) + 2 = 2\)

Thus, the intersection point is \((2, -1, 2)\).

### Step 5: Calculate the Distance
Now we calculate the distance between the point \((-1, -5, -10)\) and the intersection point \((2, -1, 2)\) using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
Substituting the coordinates:
\[ d = \sqrt{(2 - (-1))^2 + (-1 - (-5))^2 + (2 - (-10))^2} \]
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 intersection point is \(13\).