If the distances of the point (1 , 2, a) from the line $$\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}$$ along the lines $$L_{1}:\frac{x-1}{3}=\frac{y-2}{4}=\frac{z-a}{b}$$ and $$L_{2}:\frac{x-1}{1}=\frac{y-2}{4}=\frac{z-a}{c}$$ are equal, then a + b + c is equal to

We wish to determine the sum $$a + b + c$$ under the condition that the distances from the point $$P = (1,2,a)$$ to the line $$\frac{x-1}{1} = \frac{y}{2} = \frac{z-1}{1}$$ along two different directions are equal.

The given line passes through $$(1,0,1)$$ and has direction vector $$(1,2,1)$$. Let $$L_{1}$$ be the line through $$P$$ with direction $$(3,4,b)$$, and let $$L_{2}$$ be the line through $$P$$ with direction $$(1,4,c)$$.

Points on $$L_{1}$$ can be written as $$(1+3t,\;2+4t,\;a+bt)$$, while points on the given line are $$(1+s,\;2s,\;1+s)$$. Equating coordinates gives:
$$1 + 3t = 1 + s$$
$$2 + 4t = 2s$$
$$a + bt = 1 + s\,. $$

From $$1+3t=1+s$$ we have $$s=3t$$, and from $$2+4t=2s=6t$$ it follows that $$2+4t=6t\implies t=1,\;s=3$$. Substituting into the third equation yields $$a + b = 1 + 3 = 4\,. \quad(\ast)$$

Similarly, points on $$L_{2}$$ are $$(1+t',\;2+4t',\;a+ct')$$, which must equal $$(1+s',\;2s',\;1+s')$$. Equating coordinates gives:
$$1 + t' = 1 + s'$$
$$2 + 4t' = 2s'$$
$$a + ct' = 1 + s'\,. $$

From $$1+t'=1+s'$$ we get $$s'=t'$$, and from $$2+4t'=2t'$$ we obtain $$2+4t'=2t'\implies t'=-1,\;s'=-1$$. Substitution into the third equation gives $$a - c = 1 - 1 = 0\implies a=c\,. \quad(\ast\ast)$$

The intersection of $$L_{1}$$ with the given line occurs at $$t=1$$, namely at $$(4,6,4)$$, so the distance from $$P=(1,2,a)$$ is $$d_{1} = \sqrt{9 + 16 + (4 - a)^{2}}\,. $$ The intersection of $$L_{2}$$ occurs at $$t'=-1$$, namely at $$(0,-2,0)$$, giving $$d_{2} = \sqrt{1 + 16 + a^{2}}\,. $$

Setting $$d_{1}=d_{2}$$ leads to $$9 + 16 + (4 - a)^{2} = 1 + 16 + a^{2},$$ that is $$25 + 16 - 8a + a^{2} = 17 + a^{2}\implies 41 - 8a = 17\implies 8a = 24\implies a = 3\,. $$

From $$(\ast)$$ we get $$b = 4 - a = 1$$, and from $$(\ast\ast)$$ we have $$c = a = 3$$. Therefore $$a + b + c = 3 + 1 + 3 = 7\,. $$

Hence, the required sum is 7.