Let the lines `L_1: vecr=lamda(hati+2hatj+3hatk), lamdainR`
`L_2: vecr=(hati+3hatj+hatk)+mu(hati+hatj+5hatk),muinR`
intersect at the point S. If a plane `ax + by – z+d=0` passes through S and is parallel to both the lines `L_1 and L_2` then the value of a + b + d is equal to ______

Answer

To solve the problem, we will follow these steps:

### Step 1: Find the point of intersection \( S \) of the lines \( L_1 \) and \( L_2 \).

The equations of the lines are given as:
- \( L_1: \vec{r} = \lambda (\hat{i} + 2\hat{j} + 3\hat{k}) \)
- \( L_2: \vec{r} = (\hat{i} + 3\hat{j} + \hat{k}) + \mu (\hat{i} + \hat{j} + 5\hat{k}) \)

We can express \( L_1 \) and \( L_2 \) in parametric form:
- For \( L_1 \):
- \( x_1 = \lambda \)
- \( y_1 = 2\lambda \)
- \( z_1 = 3\lambda \)

- For \( L_2 \):
- \( x_2 = 1 + \mu \)
- \( y_2 = 3 + \mu \)
- \( z_2 = 1 + 5\mu \)

To find the intersection, we set \( x_1 = x_2 \), \( y_1 = y_2 \), and \( z_1 = z_2 \):
1. \( \lambda = 1 + \mu \) (1)
2. \( 2\lambda = 3 + \mu \) (2)
3. \( 3\lambda = 1 + 5\mu \) (3)

### Step 2: Solve the equations simultaneously.

From equation (1):
\[ \mu = \lambda - 1 \]

Substituting \( \mu \) into equation (2):
\[ 2\lambda = 3 + (\lambda - 1) \\ 2\lambda = 3 + \lambda - 1 \\ 2\lambda - \lambda = 2 \\ \lambda = 2 \]

Now substituting \( \lambda = 2 \) back into equation (1):
\[ \mu = 2 - 1 = 1 \]

### Step 3: Find the coordinates of the intersection point \( S \).

Using \( \lambda = 2 \) in \( L_1 \):
\[ S = (2, 4, 6) \]

### Step 4: Determine the direction vectors of the lines.

The direction vector of \( L_1 \) is \( \hat{i} + 2\hat{j} + 3\hat{k} \) and for \( L_2 \) is \( \hat{i} + \hat{j} + 5\hat{k} \).

### Step 5: Find the normal vector of the plane.

The plane is parallel to both lines, so its normal vector can be found using the cross product of the direction vectors of \( L_1 \) and \( L_2 \).

Let:
\[ \vec{d_1} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}, \quad \vec{d_2} = \begin{pmatrix} 1 \\ 1 \\ 5 \end{pmatrix} \]

Calculating the cross product \( \vec{d_1} \times \vec{d_2} \):
\[ \vec{n} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \times \begin{pmatrix} 1 \\ 1 \\ 5 \end{pmatrix} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 1 & 1 & 5 \end{vmatrix} \]

Calculating the determinant:
\[ \vec{n} = \hat{i}(2 \cdot 5 - 3 \cdot 1) - \hat{j}(1 \cdot 5 - 3 \cdot 1) + \hat{k}(1 \cdot 1 - 2 \cdot 1) \\ = \hat{i}(10 - 3) - \hat{j}(5 - 3) + \hat{k}(1 - 2) \\ = 7\hat{i} - 2\hat{j} - 1\hat{k} \]

Thus, the normal vector \( \vec{n} = (7, -2, -1) \).

### Step 6: Write the equation of the plane.

The equation of the plane can be written as:
\[ 7(x - 2) - 2(y - 4) - 1(z - 6) = 0 \]

Expanding this:
\[ 7x - 14 - 2y + 8 - z + 6 = 0 \\ 7x - 2y - z = 0 \]

This can be rewritten as:
\[ 7x - 2y - z + 0 = 0 \]

### Step 7: Identify coefficients \( a, b, d \).

From the plane equation \( ax + by - z + d = 0 \):
- \( a = 7 \)
- \( b = -2 \)
- \( d = 0 \)

### Step 8: Calculate \( a + b + d \).

\[ a + b + d = 7 - 2 + 0 = 5 \]

Thus, the value of \( a + b + d \) is \( \boxed{5} \).

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