Find the distance of the point `(2, 3, 4)` from the line `(x+3)/3=(y-2)/6=z/2` measured parallel to the plane `3x + 2y + 2z - 5= 0`.

Options

A. 2
B. 4
C. 6
D. 7

Explanation

To find the distance of the point \( P(2, 3, 4) \) from the line given by the equations \( \frac{x+3}{3} = \frac{y-2}{6} = \frac{z}{2} \) measured parallel to the plane \( 3x + 2y + 2z - 5 = 0 \), we can follow these steps:

### Step 1: Identify the direction ratios of the line
The line is given in symmetric form. We can express it in parametric form:
- Let \( \frac{x + 3}{3} = \frac{y - 2}{6} = \frac{z}{2} = \lambda \).
- Then, we can write:
\[ x = 3\lambda - 3, \quad y = 6\lambda + 2, \quad z = 2\lambda \]
The direction ratios of the line are \( (3, 6, 2) \).

### Step 2: Find the normal vector of the plane
The equation of the plane is given as \( 3x + 2y + 2z - 5 = 0 \). The normal vector \( \mathbf{n} \) of the plane is:
\[ \mathbf{n} = (3, 2, 2) \]

### Step 3: Find the equation of the plane parallel to the given plane and passing through point \( P(2, 3, 4) \)
Since the required plane is parallel to the given plane, it will have the same normal vector. The equation of the plane can be expressed as:
\[ 3(x - 2) + 2(y - 3) + 2(z - 4) = 0 \]
Expanding this:
\[ 3x - 6 + 2y - 6 + 2z - 8 = 0 \implies 3x + 2y + 2z = 20 \]

### Step 4: Find the point of intersection of the line and the new plane
Substituting the parametric equations of the line into the plane equation:
\[ 3(3\lambda - 3) + 2(6\lambda + 2) + 2(2\lambda) = 20 \]
Expanding this:
\[ 9\lambda - 9 + 12\lambda + 4 + 4\lambda = 20 \]
Combining like terms:
\[ 25\lambda - 5 = 20 \implies 25\lambda = 25 \implies \lambda = 1 \]

### Step 5: Find the coordinates of the intersection point \( Q \)
Substituting \( \lambda = 1 \) back into the parametric equations:
\[ x = 3(1) - 3 = 0, \quad y = 6(1) + 2 = 8, \quad z = 2(1) = 2 \]
Thus, the coordinates of point \( Q \) are \( (0, 8, 2) \).

### Step 6: Calculate the distance \( PQ \)
Using the distance formula:
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]
where \( P(2, 3, 4) \) and \( Q(0, 8, 2) \):
\[ d = \sqrt{(0 - 2)^2 + (8 - 3)^2 + (2 - 4)^2} \]
Calculating each term:
\[ = \sqrt{(-2)^2 + (5)^2 + (-2)^2} = \sqrt{4 + 25 + 4} = \sqrt{33} \]

### Final Answer
The distance of the point \( (2, 3, 4) \) from the line measured parallel to the plane is \( \sqrt{33} \).

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