The distance of the point (2, 3, 2) from the plane `3x+4y+4z=23` measured parallel to the line `(x+3)/(1)=(y-6)/(-2)=(z-1)/(1)` is

Options

A. `sqrt(108)` units
B. 12 units
C. `sqrt(54)` units
D. `sqrt(236)` units

Answer

To find the distance of the point \( P(2, 3, 2) \) from the plane \( 3x + 4y + 4z = 23 \) measured parallel to the line given by \( \frac{x+3}{1} = \frac{y-6}{-2} = \frac{z-1}{1} \), we can follow these steps:

### Step 1: Determine the Direction Ratios of the Line
The given line can be expressed in parametric form. The direction ratios of the line are \( (1, -2, 1) \).

### Step 2: Write the Parametric Equations of the Line
Since the line passes through the point \( (2, 3, 2) \) and has direction ratios \( (1, -2, 1) \), we can write the parametric equations as:
\[ x = 2 + t, \quad y = 3 - 2t, \quad z = 2 + t \]
where \( t \) is a parameter.

### Step 3: General Point on the Line
A general point on the line can be expressed as:
\[ (2 + t, 3 - 2t, 2 + t) \]

### Step 4: Substitute into the Plane Equation
To find the point where this line intersects the plane, substitute the parametric equations into the plane equation \( 3x + 4y + 4z = 23 \):
\[ 3(2 + t) + 4(3 - 2t) + 4(2 + t) = 23 \]

### Step 5: Simplify the Equation
Expanding and simplifying:
\[ 6 + 3t + 12 - 8t + 8 + 4t = 23 \]
Combine like terms:
\[ (3t - 8t + 4t) + (6 + 12 + 8) = 23 \]
\[ - t + 26 = 23 \]

### Step 6: Solve for \( t \)
Rearranging gives:
\[ -t = 23 - 26 \implies -t = -3 \implies t = 3 \]

### Step 7: Find the Point of Intersection
Substituting \( t = 3 \) back into the parametric equations:
\[ x = 2 + 3 = 5, \quad y = 3 - 2(3) = -3, \quad z = 2 + 3 = 5 \]
Thus, the point of intersection is \( (5, -3, 5) \).

### Step 8: Calculate the Distance
Now, we calculate the distance from the point \( P(2, 3, 2) \) to the point of intersection \( (5, -3, 5) \):
\[ \text{Distance} = \sqrt{(5 - 2)^2 + (-3 - 3)^2 + (5 - 2)^2} \]
Calculating each term:
\[ = \sqrt{(3)^2 + (-6)^2 + (3)^2} \]
\[ = \sqrt{9 + 36 + 9} = \sqrt{54} = 3\sqrt{6} \]

### Final Answer
Thus, the distance of the point \( (2, 3, 2) \) from the plane \( 3x + 4y + 4z = 23 \) measured parallel to the given line is \( 3\sqrt{6} \).