Let the lines $$L_1 : \vec{r} = \lambda(\hat{i} + 2\hat{j} + 3\hat{k}), \lambda \in R$$ and $$L_2 : \vec{r} = \hat{i} + 3\hat{j} + \hat{k} + \mu(\hat{i} + \hat{j} + 5\hat{k}); \mu \in R$$, intersect at the point $$S$$. If a plane $$ax + by - z + d = 0$$ passes through $$S$$ and is parallel to the lines $$L_1$$ and $$L_2$$, then the value of $$a + b + d$$ is equal to ______.

Given,

$$L_1:\vec r=\lambda(\hat i+2\hat j+3\hat k)$$

and

$$L_2:\vec r=\hat i+3\hat j+\hat k+\mu(\hat i+\hat j+5\hat k)$$

Direction vectors are

$$\vec d_1=(1,2,3),\qquad \vec d_2=(1,1,5)$$

A point on $$L_1$$ is

$$(\lambda,2\lambda,3\lambda)$$

and a point on $$L_2$$ is

$$(1+\mu,3+\mu,1+5\mu)$$

Since the lines intersect,

$$\lambda=1+\mu,\qquad 2\lambda=3+\mu,\qquad 3\lambda=1+5\mu$$

Using

$$\lambda=1+\mu$$

in

$$2\lambda=3+\mu$$

we get

$$2(1+\mu)=3+\mu$$

$$\mu=1$$

Hence,

$$\lambda=2$$

Thus, the point of intersection is

$$S=(2,4,6)$$

Now, the plane

$$ax+by-z+d=0$$

has normal vector

$$\vec n=(a,b,-1)$$

Since the plane is parallel to both lines,

$$\vec n\cdot\vec d_1=0$$

and

$$\vec n\cdot\vec d_2=0$$

Therefore,

$$a+2b-3=0$$

$$a+b-5=0$$

Subtracting,

$$b=-2$$

Hence,

$$a=7$$

So, the plane becomes

$$7x-2y-z+d=0$$

Since it passes through

$$S=(2,4,6)$$

$$7(2)-2(4)-6+d=0$$

$$14-8-6+d=0$$

$$d=0$$

Therefore,

$$a+b+d=7-2+0=5$$

Hence, $$\boxed{5}$$.