The distance between the point (3,4,5) and the point where the line `(x-3)/1=(y-4)/2=(z-5)/2` meets the plane `x+y+z=17` is

Options

A. 1
B. 2
C. 3
D. none of these

Explanation

To solve the problem, we need to find the distance between the point \( P(3, 4, 5) \) and the point where the line intersects the plane given by the equation \( x + y + z = 17 \).

### Step 1: Parametrize the Line
The line is given in the symmetric form:
\[ \frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z - 5}{2} = \lambda \]
From this, we can express the coordinates of points on the line in terms of \( \lambda \):
\[ x = 3 + \lambda, \quad y = 4 + 2\lambda, \quad z = 5 + 2\lambda \]

### Step 2: Substitute into the Plane Equation
We need to find the point where this line intersects the plane \( x + y + z = 17 \). Substitute the expressions for \( x \), \( y \), and \( z \) into the plane equation:
\[ (3 + \lambda) + (4 + 2\lambda) + (5 + 2\lambda) = 17 \]
Simplifying this gives:
\[ 3 + \lambda + 4 + 2\lambda + 5 + 2\lambda = 17 \]
\[ 5\lambda + 12 = 17 \]

### Step 3: Solve for \( \lambda \)
Now, solve for \( \lambda \):
\[ 5\lambda = 17 - 12 \]
\[ 5\lambda = 5 \implies \lambda = 1 \]

### Step 4: Find the Intersection Point
Now substitute \( \lambda = 1 \) back into the equations for \( x \), \( y \), and \( z \):
\[ x = 3 + 1 = 4, \quad y = 4 + 2(1) = 6, \quad z = 5 + 2(1) = 7 \]
Thus, the point of intersection \( Q \) is \( (4, 6, 7) \).

### Step 5: Calculate the Distance
Now we need to find the distance between the points \( P(3, 4, 5) \) and \( Q(4, 6, 7) \) using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
Substituting the coordinates of \( P \) and \( Q \):
\[ d = \sqrt{(4 - 3)^2 + (6 - 4)^2 + (7 - 5)^2} \]
\[ d = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3 \]

### Final Answer
The distance between the point \( (3, 4, 5) \) and the point where the line meets the plane is \( \boxed{3} \).