# The value of $k$ for which the line $\large\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}$ to lie on the plane $2x-4y+3z=6$ is ?

$(a)\:2\:\:\:\qquad\:\:(b)\:\:-2\:\:\:\qquad\:\:(c)\:\:6\:\:\:\qquad\:\:(d)\:\:No\:real\:value\:of\:k$

For any line to lie on a plane, two conditions are to be satisfied.
1. The line should be $\perp$ to the normal $\overrightarrow n$ to the plane and
$i.e.,\:\:\overrightarrow b.\overrightarrow n=0$
2. Every point on the line should satisfy the eqn. of the plane.
Here $d.r.$ of the line $\overrightarrow b=(1,1,2)$ and a point on the line = $(4,2,k)$
Normal to the plane $\overrightarrow n=(2,-4,3)$
But $\overrightarrow b.\overrightarrow n\neq 0$
$\therefore$ the line cannot lie on the plane for any real value of $k$.