The line `(x-3)/1=(y-4)/2=(z-5)/2` cuts the plane `x+y+z=17` at

Options

A. (3,4,5)
B. (4,6,7)
C. (4,5,8)
D. (8,4,5)

Explanation

To find the point of intersection of the line given by the equation \((x-3)/1 = (y-4)/2 = (z-5)/2\) and the plane given by the equation \(x + y + z = 17\), we can follow these steps:

### Step 1: Parametrize the line
The line can be expressed in parametric form. We set:
\[ \frac{x-3}{1} = \frac{y-4}{2} = \frac{z-5}{2} = \lambda \]
From this, we can express \(x\), \(y\), and \(z\) in terms of \(\lambda\):
\[ x = 1\lambda + 3 = \lambda + 3 \]
\[ y = 2\lambda + 4 \]
\[ z = 2\lambda + 5 \]

### Step 2: Substitute into the plane equation
Now, we substitute these expressions for \(x\), \(y\), and \(z\) into the plane equation \(x + y + z = 17\):
\[ (\lambda + 3) + (2\lambda + 4) + (2\lambda + 5) = 17 \]

### Step 3: Simplify the equation
Combine like terms:
\[ \lambda + 3 + 2\lambda + 4 + 2\lambda + 5 = 17 \]
This simplifies to:
\[ 5\lambda + 12 = 17 \]

### Step 4: Solve for \(\lambda\)
Now, we solve for \(\lambda\):
\[ 5\lambda = 17 - 12 \]
\[ 5\lambda = 5 \]
\[ \lambda = 1 \]

### Step 5: Find the coordinates of the intersection point
Now that we have \(\lambda\), we can find the coordinates of the intersection point \(Q\):
\[ x = \lambda + 3 = 1 + 3 = 4 \]
\[ y = 2\lambda + 4 = 2(1) + 4 = 2 + 4 = 6 \]
\[ z = 2\lambda + 5 = 2(1) + 5 = 2 + 5 = 7 \]

Thus, the point of intersection \(Q\) is:
\[ Q(4, 6, 7) \]

### Final Answer
The line cuts the plane at the point \(Q(4, 6, 7)\).

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