This problem requires a comprehensive understanding of lines and planes in 3D geometry. We will systematically determine the intersection point of two lines, then find the normal vector of the plane that is parallel to these lines, and finally use this information to establish the plane's equation and calculate the desired sum.

1. Key Concepts and Formulas

• Equation of a Line: A line passing through a point $\vec{a}$ and parallel to a direction vector $\vec{b}$ is given by $\vec{r} = \vec{a} + \lambda \vec{b}$.
• Intersection of Lines: Two lines intersect if there exist unique parameters (e.g., $\lambda$ and $\mu$) for which their position vectors are equal. This leads to a system of linear equations in $\lambda$ and $\mu$.
• Normal Vector of a Plane: If a plane is parallel to two lines with direction vectors $\vec{b_1}$ and $\vec{b_2}$, its normal vector $\vec{n}$ can be found by their cross product: $\vec{n} = \vec{b_1} \times \vec{b_2}$.
• Equation of a Plane: The equation of a plane passing through a point $(x_0, y_0, z_0)$ with a normal vector $\vec{n} = A\widehat{i} + B\widehat{j} + C\widehat{k}$ is $A(x-x_0) + B(y-y_0) + C(z-z_0) = 0$, which simplifies to $Ax + By + Cz + D = 0$.

2. Step-by-Step Solution

Step 1: Find the Intersection Point S of Lines $L_1$ and $L_2$

What we are doing: We are finding the coordinates of the point S where the two given lines meet. Why this step: The problem states that the plane passes through this point S. Therefore, determining S is essential to write the equation of the plane.

The equations of the lines are: $L_1:\overrightarrow r = \lambda \left( {\widehat i + 2\widehat j + 3\widehat k} \right) = \lambda\widehat i + 2\lambda\widehat j + 3\lambda\widehat k$ $L_2:\overrightarrow r = \left( {\widehat i + 3\widehat j + \widehat k} \right) + \mu \left( {\widehat i + \widehat j + 5\widehat k} \right) = (1+\mu)\widehat i + (3+\mu)\widehat j + (1+5\mu)\widehat k$

For the lines to intersect at point S, their corresponding coordinates must be equal for some specific values of $\lambda$ and $\mu$. Equating the components:

1. x-component: $\lambda = 1 + \mu$ (Equation 1)
2. y-component: $2\lambda = 3 + \mu$ (Equation 2)
3. z-component: $3\lambda = 1 + 5\mu$ (Equation 3)

Now, we solve this system of equations for $\lambda$ and $\mu$. Substitute Equation 1 into Equation 2: $2(1 + \mu) = 3 + \mu$ $2 + 2\mu = 3 + \mu$ $\mu = 1$

Substitute $\mu = 1$ back into Equation 1 to find $\lambda$: $\lambda = 1 + 1$ $\lambda = 2$

Verification: To ensure these values are correct, we check if they satisfy Equation 3: $3\lambda = 1 + 5\mu$ $3(2) = 1 + 5(1)$ $6 = 1 + 5$ $6 = 6$ The values $\lambda=2$ and $\mu=1$ are consistent.

Now, substitute $\lambda = 2$ into the equation for $L_1$ to find the coordinates of S: $\overrightarrow r_S = 2(\widehat i + 2\widehat j + 3\widehat k) = 2\widehat i + 4\widehat j + 6\widehat k$ Thus, the intersection point is $S = (2, 4, 6)$.

Step 2: Determine the Plane's Normal Vector

What we are doing: We are finding a vector perpendicular to the plane, which is crucial for defining its equation. Why this step: The problem states the plane is parallel to both $L_1$ and $L_2$. This means the plane's normal vector must be perpendicular to the direction vectors of both lines. The cross product provides such a vector.

The direction vectors of the lines are: For $L_1$: $\vec{b_1} = \widehat i + 2\widehat j + 3\widehat k$ For $L_2$: $\vec{b_2} = \widehat i + \widehat j + 5\widehat k$

The normal vector $\vec{n}$ to the plane is given by the cross product of $\vec{b_1}$ and $\vec{b_2}$: $\vec{n} = \vec{b_1} \times \vec{b_2}$ $\vec{n} = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ 1 & 2 & 3 \\ 1 & 1 & 5 \end{vmatrix}$ $\vec{n} = \widehat i (2 \cdot 5 - 3 \cdot 1) - \widehat j (1 \cdot 5 - 3 \cdot 1) + \widehat k (1 \cdot 1 - 2 \cdot 1)$ $\vec{n} = \widehat i (10 - 3) - \widehat j (5 - 3) + \widehat k (1 - 2)$ $\vec{n} = 7\widehat i - 2\widehat j - \widehat k$ So, a normal vector to the plane is $\vec{n} = 7\widehat i - 2\widehat j - \widehat k$.

Step 3: Construct the Plane Equation and Find $a, b, d$

What we are doing: We are using the normal vector and the intersection point S to determine the specific coefficients $a, b, d$ of the plane equation. Why this step: The problem asks for the value of $a+b+d$, so we need to find each of these coefficients.

The given form of the plane equation is $ax + by - z + d = 0$. The normal vector derived from this equation is $\vec{n}_{plane} = a\widehat{i} + b\widehat{j} - 1\widehat{k}$.

Since $\vec{n}_{plane}$ is parallel to our calculated normal vector $\vec{n}$, they must be scalar multiples of each other. $\vec{n}_{plane} = k \cdot \vec{n}$ $a\widehat{i} + b\widehat{j} - \widehat{k} = k (7\widehat{i} - 2\widehat{j} - \widehat{k})$

Comparing the components:

• $\widehat k$-component: $-1 = k(-1) \implies k = 1$
• $\widehat i$-component: $a = k(7) \implies a = 1 \cdot 7 \implies a = 7$
• $\widehat j$-component: $b = k(-2) \implies b = 1 \cdot (-2) \implies b = -2$

Now we have $a=7$ and $b=-2$. The plane equation becomes $7x - 2y - z + d = 0$.

The plane passes through the point $S=(2, 4, 6)$. Substitute these coordinates into the plane equation to find $d$: $7(2) - 2(4) - (6) + d = 0$ $14 - 8 - 6 + d = 0$ $0 + d = 0$ $d = 0$

So, the equation of the plane is $7x - 2y - z = 0$.

Step 4: Calculate $a+b+d$

What we are doing: We are summing the values of $a, b,$ and $d$ that we found. Why this step: This is the final value requested by the problem.

We have: $a = 7$ $b = -2$ $d = 0$

Therefore, $a + b + d = 7 + (-2) + 0 = 5$.

3. Common Mistakes & Tips

• Sign Errors in Cross Product: Be very careful with the signs when calculating the determinant for the cross product. A common mistake is forgetting the negative sign for the $\widehat j$ component.
• Incorrectly Equating Normal Vectors: When comparing the calculated normal vector $\vec{n}$ with the plane's given normal vector $\vec{n}_{plane}$, ensure you correctly identify the scalar multiple $k$. Pay attention to the coefficients in the given plane equation (e.g., $ax+by-z+d=0$ implies a $z$-component of $-1$, not $+1$).
• Arithmetic Mistakes: Double-check all arithmetic, especially when solving the system of equations for $\lambda$ and $\mu$, and when substituting the point S into the plane equation.

4. Summary

We began by finding the intersection point S of the two given lines by equating their parametric forms, yielding $S=(2,4,6)$. Next, we determined the normal vector of the plane by taking the cross product of the direction vectors of the two lines, as the plane is parallel to both; this gave $\vec{n} = 7\widehat i - 2\widehat j - \widehat k$. By comparing this with the general normal vector of the plane $ax+by-z+d=0$, we found $a=7$ and $b=-2$. Finally, we used the point S to find $d=0$. The sum $a+b+d$ was then calculated as $7+(-2)+0 = 5$.

The final answer is $\boxed{5}$.