Find the distance from the point (3,4,5) to the point, where the line:
`(x - 3)/(1) = (y - 4)/(2) = (z - 5)/(2)`
meets the plane x + y + z = 2.

Answer

To find the distance from the point \( (3, 4, 5) \) to the point where the line intersects the plane \( x + y + z = 2 \), we will follow these steps:

### Step 1: Write the parametric equations of the line
The line is given by the equation:
\[ \frac{x - 3}{1} = \frac{y - 4}{2} = \frac{z - 5}{2} \]
Let \( t \) be the parameter. Then, we can express the coordinates \( x, y, z \) in terms of \( t \):
\[ x = 3 + t \]
\[ y = 4 + 2t \]
\[ z = 5 + 2t \]

### Step 2: Substitute the parametric equations into the plane equation
The plane equation is:
\[ x + y + z = 2 \]
Substituting the parametric equations into the plane equation gives:
\[ (3 + t) + (4 + 2t) + (5 + 2t) = 2 \]
Simplifying this:
\[ 3 + t + 4 + 2t + 5 + 2t = 2 \]
\[ 3 + 4 + 5 + t + 2t + 2t = 2 \]
\[ 12 + 5t = 2 \]

### Step 3: Solve for \( t \)
Now, we solve for \( t \):
\[ 5t = 2 - 12 \]
\[ 5t = -10 \]
\[ t = -2 \]

### Step 4: Find the coordinates of the intersection point
Now that we have \( t = -2 \), we can find the coordinates of the intersection point \( O \):
\[ x = 3 + (-2) = 1 \]
\[ y = 4 + 2(-2) = 0 \]
\[ z = 5 + 2(-2) = 1 \]
Thus, the intersection point \( O \) is \( (1, 0, 1) \).

### Step 5: Calculate the distance from the point \( (3, 4, 5) \) to the point \( (1, 0, 1) \)
The distance \( d \) between the points \( (3, 4, 5) \) and \( (1, 0, 1) \) is given by the distance formula:
\[ d = \sqrt{(3 - 1)^2 + (4 - 0)^2 + (5 - 1)^2} \]
Calculating this:
\[ d = \sqrt{(2)^2 + (4)^2 + (4)^2} \]
\[ d = \sqrt{4 + 16 + 16} \]
\[ d = \sqrt{36} = 6 \]

### Final Answer
The distance from the point \( (3, 4, 5) \) to the point where the line meets the plane is \( 6 \) units.

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