Let the foot of perpendicular from a point $$P(1, 2, -1)$$ to the straight line $$L: \frac{x}{1} = \frac{y}{0} = \frac{z}{-1}$$ be $$N$$. Let a line be drawn from $$P$$ parallel to the plane $$x + y + 2z = 0$$ which meets $$L$$ at point $$Q$$. If $$\alpha$$ is the acute angle between the lines PN and PQ, then $$\cos\alpha$$ is equal to:

We have the point $$P(1,\,2,\,-1)$$ and the straight line $$L:\; \dfrac{x}{1}= \dfrac{y}{0}= \dfrac{z}{-1}.$$

Because the middle ratio is divided by zero, the symmetric form really tells us that

$$\dfrac{x}{1}= \dfrac{z}{-1}=t, \qquad y=0.$$

So every point on $$L$$ can be written as

$$\bigl(t,\,0,\,-t\bigr),$$

and the direction vector of the line is clearly $$\vec d_L=(1,\,0,\,-1).$$

Let the foot of the perpendicular from $$P$$ to the line be $$N(t,\,0,\,-t).$$ By definition, the vector $$\overrightarrow{PN}$$ is perpendicular to the direction vector of the line. The scalar product of perpendicular vectors is zero, so we write

$$\overrightarrow{PN}\cdot\vec d_L=0.$$

First we compute $$\overrightarrow{PN}:$$

$$\overrightarrow{PN}=\bigl(t-1,\;0-2,\;-t-(-1)\bigr)=\bigl(t-1,\,-2,\,-t+1\bigr).$$

Taking the dot-product, we get

$$\bigl(t-1,\,-2,\,-t+1\bigr)\cdot(1,\,0,\,-1) =(t-1)\cdot1+(-2)\cdot0+(-t+1)\cdot(-1).$$

This simplifies to

$$\;(t-1)+0+(t-1)=2(t-1).$$

Setting it equal to zero gives

$$2(t-1)=0\;\Longrightarrow\;t=1.$$

Hence

$$N=\bigl(1,\,0,\,-1\bigr).$$

Now we compute the vector $$\overrightarrow{PN}:$$

$$\overrightarrow{PN}=N-P=(1-1,\;0-2,\,-1-(-1))=(0,\,-2,\,0).$$

Let a second point $$Q(s,\,0,\,-s)$$ on the line be such that the segment $$PQ$$ is parallel to the plane $$x+y+2z=0.$$ A line is parallel to a plane precisely when its direction vector is perpendicular to the normal vector of the plane. The normal vector of the given plane is $$\vec n=(1,\,1,\,2).$$ Therefore, for the direction vector $$\overrightarrow{PQ}$$ we must have

$$\overrightarrow{PQ}\cdot\vec n=0.$$

First we find $$\overrightarrow{PQ}:$$

$$\overrightarrow{PQ}=Q-P=(s-1,\;0-2,\,-s-(-1))=(s-1,\,-2,\,-s+1).$$

Taking the dot-product with $$\vec n=(1,\,1,\,2)$$ and setting it to zero, we write

$$\bigl(s-1,\,-2,\,-s+1\bigr)\cdot(1,\,1,\,2)=0.$$

That is

$$(s-1)\cdot1+(-2)\cdot1+(-s+1)\cdot2=0,$$

which simplifies step by step:

$$s-1-2+2(-s+1)=0,$$

$$s-1-2-2s+2=0,$$

$$(-s)-1=0,$$

$$s=-1.$$

Thus

$$Q=\bigl(-1,\,0,\,1\bigr).$$

Now we have

$$\overrightarrow{PQ}=(-1-1,\;0-2,\;1-(-1))=(-2,\,-2,\,2).$$

We are asked for the acute angle $$\alpha$$ between the two lines $$PN$$ and $$PQ,$$ that is, the angle between the vectors $$\overrightarrow{PN}=(0,\,-2,\,0)$$ and $$\overrightarrow{PQ}=(-2,\,-2,\,2).$$ By the definition of the cosine of the angle between vectors,

$$\cos\alpha=\dfrac{\overrightarrow{PN}\cdot\overrightarrow{PQ}}{\lvert\overrightarrow{PN}\rvert\,\lvert\overrightarrow{PQ}\rvert}.$$

First the dot-product:

$$\overrightarrow{PN}\cdot\overrightarrow{PQ} =(0)(-2)+(-2)(-2)+(0)(2)=0+4+0=4.$$

Then the magnitudes:

$$\lvert\overrightarrow{PN}\rvert =\sqrt{0^{2}+(-2)^{2}+0^{2}} =\sqrt{4} =2,$$

$$\lvert\overrightarrow{PQ}\rvert =\sqrt{(-2)^{2}+(-2)^{2}+2^{2}} =\sqrt{4+4+4} =\sqrt{12} =2\sqrt3.$$

Substituting into the cosine formula, we get

$$\cos\alpha =\dfrac{4}{2\cdot2\sqrt3} =\dfrac{4}{4\sqrt3} =\dfrac{1}{\sqrt3}.$$

This value is positive, so it indeed corresponds to the acute angle between the lines.

Hence, the correct answer is Option C.