If the distance between the plane 23x-10y-2z+48 =0 and the plane containing the lines
`(x+1)/2 =(y-3)/4= (z+1)/(3)` and `(x+3)/(2)=(y+2)/6 = (z-1)/(lambda)(lambda in R)` is equal to `k/sqrt(633)` , then k is equal to __________.

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to find the distance between the given plane and the plane containing the lines. Let's break down the solution step by step.

### Step 1: Identify the given plane and its equation
The equation of the first plane is given as:
\[ 23x - 10y - 2z + 48 = 0 \]
From this equation, we can identify the coefficients:
- \( A_1 = 23 \)
- \( B_1 = -10 \)
- \( C_1 = -2 \)
- \( D_1 = 48 \)

### Step 2: Identify points on the lines
We have two lines given in symmetric form. We need to find points on these lines.

**Line 1:**
\[ \frac{x + 1}{2} = \frac{y - 3}{4} = \frac{z + 1}{3} \]
Let \( t \) be the parameter. Then we can express the coordinates as:
- \( x = 2t - 1 \)
- \( y = 4t + 3 \)
- \( z = 3t - 1 \)

**Line 2:**
\[ \frac{x + 3}{2} = \frac{y + 2}{6} = \frac{z - 1}{\lambda} \]
Let \( s \) be the parameter. Then we can express the coordinates as:
- \( x = 2s - 3 \)
- \( y = 6s - 2 \)
- \( z = \lambda s + 1 \)

### Step 3: Find a point on the first line
To find a specific point on the first line, we can set \( t = 0 \):
- \( x_1 = 2(0) - 1 = -1 \)
- \( y_1 = 4(0) + 3 = 3 \)
- \( z_1 = 3(0) - 1 = -1 \)

Thus, a point on the first line is \( P_1(-1, 3, -1) \).

### Step 4: Find the distance from the point to the plane
The distance \( d \) from a point \( (x_0, y_0, z_0) \) to the plane \( Ax + By + Cz + D = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \]
Substituting the values:
- \( A = 23, B = -10, C = -2, D = 48 \)
- \( (x_0, y_0, z_0) = (-1, 3, -1) \)

Calculating the numerator:
\[ |23(-1) - 10(3) - 2(-1) + 48| = |-23 - 30 + 2 + 48| = | -23 - 30 + 2 + 48| = | -3 | = 3 \]

Calculating the denominator:
\[ \sqrt{23^2 + (-10)^2 + (-2)^2} = \sqrt{529 + 100 + 4} = \sqrt{633} \]

Thus, the distance \( d \) is:
\[ d = \frac{3}{\sqrt{633}} \]

### Step 5: Relate the distance to the given expression
We know from the problem statement that the distance is equal to \( \frac{k}{\sqrt{633}} \). Therefore, we can equate:
\[ \frac{3}{\sqrt{633}} = \frac{k}{\sqrt{633}} \]
This implies that:
\[ k = 3 \]

### Final Answer
Thus, the value of \( k \) is:
\[ \boxed{3} \]