The disatance of the point `(1, 0, 2)` from the point of intersection of the line `(x-2)/(3)=(y+1)/(4)=(z-2)/(12)` and the plane `x-y+z=16`, is

Options

A. `2sqrt(14)`
B. `8`
C. `3sqrt(21)`
D. `13`

Answer

To find the distance of the point \( (1, 0, 2) \) 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 \( x - y + z = 16 \), we will follow these steps:

### Step 1: Parametrize the Line
The line can be parametrized using a parameter \( \lambda \):
\[ x = 3\lambda + 2, \quad y = 4\lambda - 1, \quad z = 12\lambda + 2 \]

### Step 2: Substitute into the Plane Equation
Next, we substitute the parametric equations of the line into the equation of the plane \( x - y + z = 16 \):
\[ (3\lambda + 2) - (4\lambda - 1) + (12\lambda + 2) = 16 \]

### Step 3: Simplify the Equation
Now, simplify the equation:
\[ 3\lambda + 2 - 4\lambda + 1 + 12\lambda + 2 = 16 \]
Combine like terms:
\[ (3\lambda - 4\lambda + 12\lambda) + (2 + 1 + 2) = 16 \]
\[ 11\lambda + 5 = 16 \]

### Step 4: Solve for \( \lambda \)
Now, solve for \( \lambda \):
\[ 11\lambda = 16 - 5 \]
\[ 11\lambda = 11 \implies \lambda = 1 \]

### Step 5: Find the Point of Intersection
Substituting \( \lambda = 1 \) back into the parametric equations to find the coordinates of the intersection point:
\[ x = 3(1) + 2 = 5, \quad y = 4(1) - 1 = 3, \quad z = 12(1) + 2 = 14 \]
Thus, the point of intersection is \( (5, 3, 14) \).

### Step 6: Calculate the Distance
Now, we calculate the distance between the points \( (1, 0, 2) \) and \( (5, 3, 14) \) using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
Substituting the coordinates:
\[ d = \sqrt{(5 - 1)^2 + (3 - 0)^2 + (14 - 2)^2} \]
Calculating each term:
\[ = \sqrt{(4)^2 + (3)^2 + (12)^2} \]
\[ = \sqrt{16 + 9 + 144} \]
\[ = \sqrt{169} \]
\[ = 13 \]

### Final Answer
The distance of the point \( (1, 0, 2) \) from the point of intersection is \( \boxed{13} \).