The length of the projection of the line segment joining the points (5, -1, 4) and (4, -1, 3) on the plane, $$x + y + z = 7$$ is:

We have two points in space, $$P(5,\,-1,\,4)$$ and $$Q(4,\,-1,\,3)$$. First we find the vector joining these two points. By subtracting the corresponding coordinates,

$$\overrightarrow{PQ}=Q-P =\bigl(4-5,\;(-1)-(-1),\;3-4\bigr) =(-1,\;0,\,-1).$$

The length of the line segment $$PQ$$ is the magnitude of this vector. Using the distance (or magnitude) formula $$|\,(a,b,c)\,|=\sqrt{a^{2}+b^{2}+c^{2}},$$ we obtain

$$|\overrightarrow{PQ}|=\sqrt{(-1)^{2}+0^{2}+(-1)^{2}} =\sqrt{1+0+1} =\sqrt{2}.$$

Next, we are asked for the length of the projection of this segment on the plane $$x+y+z=7$$. The plane’s normal vector is obtained directly from its equation $$x+y+z=7$$, giving us

$$\mathbf{n}=(1,\,1,\,1).$$

To work comfortably, we convert this normal vector into a unit normal vector. The magnitude of $$\mathbf{n}$$ is

$$|\mathbf{n}|=\sqrt{1^{2}+1^{2}+1^{2}} =\sqrt{3},$$

so the unit normal vector is

$$\widehat{\mathbf{n}} =\frac{\mathbf{n}}{|\mathbf{n}|} =\bigl(\tfrac{1}{\sqrt{3}},\;\tfrac{1}{\sqrt{3}},\;\tfrac{1}{\sqrt{3}}\bigr).$$

The component of $$\overrightarrow{PQ}$$ along this normal direction is found with the dot product formula $$\text{(component)} = (\overrightarrow{PQ}\,\cdot\,\widehat{\mathbf{n}}).$$ We compute the dot product step by step:

$$\overrightarrow{PQ}\cdot\widehat{\mathbf{n}} =(-1,0,-1)\cdot\Bigl(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\Bigr) =\frac{-1}{\sqrt{3}}+0+\frac{-1}{\sqrt{3}} =\frac{-2}{\sqrt{3}}.$$

Hence the magnitude of the component of $$\overrightarrow{PQ}$$ perpendicular to the plane is

$$\left|\,\overrightarrow{PQ}\cdot\widehat{\mathbf{n}}\,\right| =\left|\,\frac{-2}{\sqrt{3}}\,\right| =\frac{2}{\sqrt{3}}.$$

The projection of $$\overrightarrow{PQ}$$ onto the plane is what remains after removing this perpendicular component. A useful relation is

$$|\text{projection on plane}|^{2} =|\overrightarrow{PQ}|^{2} -\bigl(\overrightarrow{PQ}\cdot\widehat{\mathbf{n}}\bigr)^{2}.$$

Substituting the values we have already calculated,

$$|\text{projection on plane}|^{2} =\bigl(\sqrt{2}\bigr)^{2} -\left(\frac{-2}{\sqrt{3}}\right)^{2} =2-\frac{4}{3} =\frac{6-4}{3} =\frac{2}{3}.$$

Taking the square root, the required length is

$$|\text{projection on plane}| =\sqrt{\frac{2}{3}}.$$

Hence, the correct answer is Option A.