which of the following lines lie on the plane x+ 2y - z + 4= 0?

For line $\frac{x-1}{1}=\frac{y}{-1}=\frac{z-5}{-1}$ point (1, 0, 5) lies on 1-1-ln^n the plane .

Also, the vector along the line $\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$ is perpendicular to the normal $\hat{\mathrm{i}}+2\hat{\mathrm{j}}-\mathrm{k}$  to the plane.  For line $\overrightarrow{\mathrm{r}}=2\widehat{\mathrm{i}}-\widehat{\mathrm{j}}+4\widehat{\mathrm{k}}+\mathrm{\lambda }(3\widehat{\mathrm{i}}+\widehat{\mathrm{j}}+5\widehat{\mathrm{k}})$

 point (2, -1,4) lies on the plane and vector $3\widehat{i}+\widehat{j}+5\widehat{k}$ is perpendicular to the normal $\widehat{i}+2\widehat{j}-\widehat{k}$

Line x - y + z = 2x + y - z= 0 passes through the origin, which is not on the given plane.