If the lines `(x-1)/1=(y-2)/2=(z+3)/1` and `(x-a)/2=(y+2)/3=(z-3)/1` intersect at the point P, then the distance of the point P from the plane z = a is :

Options

A. 22
B. 10
C. 28
D. 16

Answer

To solve the problem, we need to find the point of intersection \( P \) of the two lines given by their symmetric equations and then calculate the distance of point \( P \) from the plane \( z = a \).

### Step 1: Write the equations of the lines in parametric form

The first line is given by:
\[ \frac{x-1}{1} = \frac{y-2}{2} = \frac{z+3}{1} \]
Let \( \lambda \) be the parameter. Then, we can express the coordinates as:
\[ x = \lambda + 1, \quad y = 2\lambda + 2, \quad z = \lambda - 3 \]

The second line is given by:
\[ \frac{x-a}{2} = \frac{y+2}{3} = \frac{z-3}{1} \]
Let \( \mu \) be the parameter for this line. Then, we can express the coordinates as:
\[ x = 2\mu + a, \quad y = 3\mu - 2, \quad z = \mu + 3 \]

### Step 2: Set the coordinates equal to find the intersection

To find the point of intersection, we set the coordinates equal to each other:
1. \( \lambda + 1 = 2\mu + a \) (1)
2. \( 2\lambda + 2 = 3\mu - 2 \) (2)
3. \( \lambda - 3 = \mu + 3 \) (3)

### Step 3: Solve the equations

From equation (3):
\[ \lambda - 3 = \mu + 3 \implies \lambda = \mu + 6 \tag{4} \]

Substituting \( \lambda \) from (4) into equation (1):
\[ (\mu + 6) + 1 = 2\mu + a \implies 7 - a = \mu \tag{5} \]

Substituting \( \lambda \) from (4) into equation (2):
\[ 2(\mu + 6) + 2 = 3\mu - 2 \implies 2\mu + 12 + 2 = 3\mu - 2 \]
\[ 14 + 2 = 3\mu - 2\mu \implies \mu = 16 \tag{6} \]

### Step 4: Find \( a \)

Substituting \( \mu = 16 \) into equation (5):
\[ 7 - a = 16 \implies a = 7 - 16 = -9 \]

### Step 5: Find the coordinates of point \( P \)

Substituting \( \mu = 16 \) back into the parametric equations of either line to find \( P \):
Using the first line:
\[ \lambda = \mu + 6 = 16 + 6 = 22 \]
Now substituting \( \lambda \) into the first line's equations:
\[ x = 22 + 1 = 23, \quad y = 2(22) + 2 = 46, \quad z = 22 - 3 = 19 \]
Thus, the coordinates of point \( P \) are \( (23, 46, 19) \).

### Step 6: Calculate the distance from the point \( P \) to the plane \( z = a \)

The plane is given by \( z = a = -9 \). The distance \( d \) from point \( P \) to the plane is given by:
\[ d = |z_P - a| = |19 - (-9)| = |19 + 9| = |28| = 28 \]

### Final Answer
The distance of the point \( P \) from the plane \( z = a \) is \( \boxed{28} \).