Let L be the line $$\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+3}{6}$$ and let S be the set of all points (a, b, c) on L, whose distance from the line $$\frac{x+1}{2}=\frac{y+1}{3}=\frac{z-9}{0}$$a long the line L is 7. Then $$\sum_{(a,b,c)\in S} (a+b+c) $$ is equal to :

We are given the line $$L: \frac{x+1}{2} = \frac{y+1}{3} = \frac{z+3}{6}$$, which passes through $$(-1,-1,-3)$$ with direction ratios $$(2,3,6)$$, and the second line $$\frac{x+1}{2} = \frac{y+1}{3} = \frac{z-9}{0}$$ passing through $$(-1,-1,9)$$ with direction ratios $$(2,3,0)$$. We seek points $$(a,b,c)$$ on $$L$$ whose distance from the second line, measured along $$L$$, is 7.

Parametrizing $$L$$ by $$\mathbf{r}_1 = (-1,-1,-3) + t(2,3,6)$$ shows that a general point on $$L$$ is $$(2t-1,3t-1,6t-3)\,.$$

The phrase “distance along $$L$$ from a point $$(a,b,c)$$ to the second line” means the length of the segment on $$L$$ from $$(a,b,c)$$ to the foot of the common perpendicular between the two lines. Thus we first find where $$L$$ comes closest to the second line, parametrized by $$\mathbf{r}_2 = (-1,-1,9) + s(2,3,0)\,.$$

The connecting vector is $$\mathbf{r}_1-\mathbf{r}_2 = (2t-2s,3t-3s,6t-12)\,. $$ For this to be perpendicular to both direction vectors, we impose

$$(2t-2s,3t-3s,6t-12)\cdot(2,3,6)=0$$, which gives $$4t-4s+9t-9s+36t-72=0\quad\Rightarrow\quad49t-13s=72\quad\text{(i)}\,, $$ and

$$(2t-2s,3t-3s,6t-12)\cdot(2,3,0)=0$$, which gives $$4t-4s+9t-9s=0\quad\Rightarrow\quad13t-13s=0\quad\Rightarrow\quad t=s\quad\text{(ii)}\,. $$

Substituting $$t=s$$ from (ii) into (i) yields $$49t-13t=72\;\Rightarrow\;36t=72\;\Rightarrow\;t=2\,. $$ Hence the foot of the common perpendicular on $$L$$ is $$(2\cdot2-1,\,3\cdot2-1,\,6\cdot2-3)=(3,5,9)\,.$$

Since the direction vector $$(2,3,6)$$ has magnitude $$\sqrt{4+9+36}=7$$, a unit change $$\Delta t=1$$ corresponds to a distance of 7 along $$L$$. Therefore, the points at distance 7 from $$(3,5,9)$$ along $$L$$ occur at $$t=2\pm1$$, giving:

For $$t=3$$: $$(2\cdot3-1,\,3\cdot3-1,\,6\cdot3-3)=(5,8,15)\quad\text{so}\quad a+b+c=28\,, $$

For $$t=1$$: $$(2\cdot1-1,\,3\cdot1-1,\,6\cdot1-3)=(1,2,3)\quad\text{so}\quad a+b+c=6\,. $$

Summing these values gives $$28+6=34$$, so the answer is 34, corresponding to Option 4.