if the line $$\frac {x-4}{1}=\frac{y-2}{1}=\frac{z-p}{2}$$ lies on the place $$2x-4y+z=7$$, then value of p is​

Correct option is C

​$$\textbf{Solution} \\\text{Line: } \frac{x - 4}{1} = \frac{y - 2}{1} = \frac{z - p}{2} \\\text{Plane: } 2x - 4y + z = 7 \\[8pt]\text{Let } \lambda \in \mathbb{R} \text{ be the parameter. Then:} \\x = 4 + \lambda,\quad y = 2 + \lambda,\quad z = p + 2\lambda \\[6pt]\text{Substitute into the plane equation:} \\2(4 + \lambda) - 4(2 + \lambda) + (p + 2\lambda) = 7 \\[6pt]8 + 2\lambda - 8 - 4\lambda + p + 2\lambda = 7 \\[6pt](2\lambda - 4\lambda + 2\lambda) + p = 7 \Rightarrow 0 + p = 7 \\[6pt]\Rightarrow \boxed{p = 7}$$

​​