If the line `(x-1)/(2)=(y+1)/(3)=(z-2)/(4)` meets the plane, `x+2y+3z=15` at a point P, then the distance of P from the origin is

Options

A. `9//2`
B. `sqrt(5)//2`
C. `2 sqrt(5)`
D. `7//2`

Explanation

To solve the problem step by step, we will find the point of intersection \( P \) of the given line and plane, and then calculate the distance from the origin to point \( P \).

### Step 1: Write the equations of the line and the plane.
The line is given in symmetric form:
\[ \frac{x-1}{2} = \frac{y+1}{3} = \frac{z-2}{4} \]
The plane is given by the equation:
\[ x + 2y + 3z = 15 \]

### Step 2: Parametrize the line.
Let \( t \) be the parameter. We can express the coordinates \( x, y, z \) in terms of \( t \):
\[ x = 2t + 1 \]
\[ y = 3t - 1 \]
\[ z = 4t + 2 \]

### Step 3: Substitute the parametric equations into the plane equation.
Substituting \( x, y, z \) into the plane equation:
\[ (2t + 1) + 2(3t - 1) + 3(4t + 2) = 15 \]
Expanding this:
\[ 2t + 1 + 6t - 2 + 12t + 6 = 15 \]
Combining like terms:
\[ (2t + 6t + 12t) + (1 - 2 + 6) = 15 \]
\[ 20t + 5 = 15 \]

### Step 4: Solve for \( t \).
Subtract 5 from both sides:
\[ 20t = 10 \]
Dividing by 20:
\[ t = \frac{1}{2} \]

### Step 5: Find the coordinates of point \( P \).
Substituting \( t = \frac{1}{2} \) back into the parametric equations:
\[ x = 2\left(\frac{1}{2}\right) + 1 = 1 + 1 = 2 \]
\[ y = 3\left(\frac{1}{2}\right) - 1 = \frac{3}{2} - 1 = \frac{1}{2} \]
\[ z = 4\left(\frac{1}{2}\right) + 2 = 2 + 2 = 4 \]
Thus, the coordinates of point \( P \) are \( (2, \frac{1}{2}, 4) \).

### Step 6: Calculate the distance from the origin to point \( P \).
The distance \( d \) from the origin \( (0, 0, 0) \) to point \( P(2, \frac{1}{2}, 4) \) is given by:
\[ d = \sqrt{(2 - 0)^2 + \left(\frac{1}{2} - 0\right)^2 + (4 - 0)^2} \]
Calculating each term:
\[ d = \sqrt{2^2 + \left(\frac{1}{2}\right)^2 + 4^2} \]
\[ = \sqrt{4 + \frac{1}{4} + 16} \]
\[ = \sqrt{20 + \frac{1}{4}} = \sqrt{\frac{80}{4} + \frac{1}{4}} = \sqrt{\frac{81}{4}} = \frac{9}{2} \]

### Final Answer:
The distance of point \( P \) from the origin is \( \frac{9}{2} \).

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