Projection of line segment joining (2, 3, 4) and (5, 6, 7) on plane $2\mathrm{x}+\mathrm{y}+\mathrm{z}=1$ is

Let A = (2, 3, 4) and B = (5, 6, 7).AB
Then $\mathrm{AB}=3\sqrt{3}$

Plane : 2x + y + z = 1
Projection of $\overline{\mathrm{AB}}$on vector $2\widehat{\mathrm{i}}+\widehat{\mathrm{j}}+\widehat{\mathrm{k}}$ which is normal to plane.
Normal to plane,
$$\begin{gathered} \mathrm{B}=\frac{(3\widehat{\mathrm{i}}+3\widehat{\mathrm{j}}+3\widehat{\mathrm{k}}).(2\widehat{\mathrm{i}}+\widehat{\mathrm{j}}+\widehat{\mathrm{k}})}{\sqrt{6}}=\frac{12}{\sqrt{6}}=2\sqrt{6} \\ \therefore \mathrm{Projection} \mathrm{of} \overline{\mathrm{AB}} \mathrm{on} \mathrm{plane}, \\ \mathrm{AC}=\sqrt{{\left(3\sqrt{3}\right)}^2-{\left(2\sqrt{6}\right)}^2}=\sqrt{3} \end{gathered}$$