Let `L_(1):x=y=z and L_(2)=x-1=y-2=z-3` be two lines. The foot of perpendicular drawn from the origin `O(0, 0, 0)` on `L_(1)" to "L_(2)` is A. If the equation of a plane containing the line `L_(1)` and perpendicular to OA is `10x+by+cz=d`, then the value of `b+c+d` is equal to

Options

A. 10
B. `-10`
C. 12
D. `-7`

Answer

To solve the problem, we need to find the coordinates of point A, which is the foot of the perpendicular drawn from the origin O(0, 0, 0) to line L2, and then determine the equation of the plane containing line L1 and perpendicular to OA.

### Step 1: Identify the equations of the lines
The equations of the lines are given as:
- Line L1: \( x = y = z \)
- Line L2: \( x - 1 = y - 2 = z - 3 \)

### Step 2: Parametrize the lines
We can parametrize the lines as follows:
- For line L1, we can set \( x = t, y = t, z = t \). Thus, any point on L1 can be represented as \( (t, t, t) \).
- For line L2, we can set \( x = s + 1, y = s + 2, z = s + 3 \). Thus, any point on L2 can be represented as \( (s + 1, s + 2, s + 3) \).

### Step 3: Find the coordinates of point A
Point A is the foot of the perpendicular from the origin O(0, 0, 0) to line L2. The direction vector of OA can be represented as:
\[ \text{OA} = (s + 1, s + 2, s + 3) - (0, 0, 0) = (s + 1, s + 2, s + 3) \]
The direction vector of line L1 is \( (1, 1, 1) \). For OA to be perpendicular to L1, the dot product must be zero:
\[ (s + 1) + (s + 2) + (s + 3) = 0 \]
This simplifies to:
\[ 3s + 6 = 0 \implies s = -2 \]

### Step 4: Substitute s to find coordinates of A
Substituting \( s = -2 \) into the parametrization of line L2 gives:
\[ A = (-2 + 1, -2 + 2, -2 + 3) = (-1, 0, 1) \]

### Step 5: Find the direction vector OA
The direction vector OA from the origin to point A is:
\[ \text{OA} = (-1, 0, 1) \]

### Step 6: Find the normal vector of the plane
The normal vector of the plane containing line L1 and perpendicular to OA is given by the cross product of the direction vector of L1 and OA. The direction vector of L1 is \( (1, 1, 1) \).

Calculating the cross product:
\[ \text{n} = (1, 1, 1) \times (-1, 0, 1) \]
Using the determinant formula:
\[ \text{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ -1 & 0 & 1 \end{vmatrix} = \hat{i}(1 \cdot 1 - 1 \cdot 0) - \hat{j}(1 \cdot 1 - 1 \cdot -1) + \hat{k}(1 \cdot 0 - 1 \cdot -1) \]
This results in:
\[ \text{n} = \hat{i}(1) - \hat{j}(2) + \hat{k}(1) = (1, -2, 1) \]

### Step 7: Equation of the plane
The equation of the plane can be expressed as:
\[ 1(x - 0) - 2(y - 0) + 1(z - 0) = 0 \]
This simplifies to:
\[ x - 2y + z = 0 \]

### Step 8: Compare with the given plane equation
The equation of the plane is given in the form:
\[ 10x + by + cz = d \]
From our derived equation \( x - 2y + z = 0 \), we can rewrite it as:
\[ 10x + 0y + 0z = 0 \]
This indicates:
- \( b = 0 \)
- \( c = -2 \)
- \( d = 0 \)

### Step 9: Calculate \( b + c + d \)
Now, we compute:
\[ b + c + d = 0 - 2 + 0 = -2 \]

Thus, the final answer is:
\[ \boxed{-2} \]