The distance of the point (–1, 9,–16) from the plane 2x + 3y – z = 5 measured parallel to the line `(x+4)/3=(2-y)/4=(z-3)/12` is

Options

A. `20sqrt2`
B. 31
C. `13sqrt2`
D. 26

Answer

To solve the problem of finding the distance from the point \((-1, 9, -16)\) to the plane \(2x + 3y - z = 5\) measured parallel to the line given by \(\frac{x + 4}{3} = \frac{2 - y}{4} = \frac{z - 3}{12}\), we can follow these steps:

### Step 1: Identify the direction ratios of the line
The line is given in symmetric form. From the equation \(\frac{x + 4}{3} = \frac{2 - y}{4} = \frac{z - 3}{12}\), we can extract the direction ratios of the line:
- \(a = 3\)
- \(b = -4\)
- \(c = 12\)

### Step 2: Find a point on the line
To find a point on the line, we can set the parameter \(t = 0\):
\[ x + 4 = 3t \implies x = 3t - 4 \quad \text{(for } t = 0 \text{, } x = -4\text{)} \]
\[ 2 - y = 4t \implies y = 2 - 4t \quad \text{(for } t = 0 \text{, } y = 2\text{)} \]
\[ z - 3 = 12t \implies z = 12t + 3 \quad \text{(for } t = 0 \text{, } z = 3\text{)} \]
Thus, a point on the line when \(t = 0\) is \((-4, 2, 3)\).

### Step 3: Find the equation of the line passing through the point \((-1, 9, -16)\) and parallel to the given line
The line parallel to the given line and passing through the point \((-1, 9, -16)\) can be expressed as:
\[ \frac{x + 1}{3} = \frac{y - 9}{-4} = \frac{z + 16}{12} \]

### Step 4: Find the parametric equations of the new line
Let \(r\) be the parameter for the new line:
\[ x = 3r - 1 \]
\[ y = -4r + 9 \]
\[ z = 12r - 16 \]

### Step 5: Substitute the parametric equations into the plane equation
Substituting \(x\), \(y\), and \(z\) into the plane equation \(2x + 3y - z = 5\):
\[ 2(3r - 1) + 3(-4r + 9) - (12r - 16) = 5 \]
Expanding this:
\[ 6r - 2 - 12r + 27 - 12r + 16 = 5 \]
Combining like terms:
\[ -18r + 41 = 5 \]
Solving for \(r\):
\[ -18r = 5 - 41 \implies -18r = -36 \implies r = 2 \]

### Step 6: Find the coordinates of the point on the plane
Substituting \(r = 2\) back into the parametric equations:
\[ x = 3(2) - 1 = 6 - 1 = 5 \]
\[ y = -4(2) + 9 = -8 + 9 = 1 \]
\[ z = 12(2) - 16 = 24 - 16 = 8 \]
Thus, the point on the plane is \((5, 1, 8)\).

### Step 7: Calculate the distance between the two points
Now, we will find the distance between the point \((-1, 9, -16)\) and the point \((5, 1, 8)\):
\[ d = \sqrt{(5 - (-1))^2 + (1 - 9)^2 + (8 - (-16))^2} \]
Calculating each component:
\[ = \sqrt{(5 + 1)^2 + (1 - 9)^2 + (8 + 16)^2} \]
\[ = \sqrt{6^2 + (-8)^2 + 24^2} \]
\[ = \sqrt{36 + 64 + 576} \]
\[ = \sqrt{676} = 26 \]

### Final Answer
The distance from the point \((-1, 9, -16)\) to the plane \(2x + 3y - z = 5\) measured parallel to the line is \(26\).