3D Coordinate Geometry -1

$\large L\equiv (x-4) = (y-2) = \dfrac {z-k}{2}$ $\large P\equiv 2x-4y+z=7$ $(4,2,k)$ is a point on the line $L$ and since whole line lies on plane $P$ , this point must also lie on $P$ . $\therefore \Large 2(4)-4(2)+1(k)=7$ $\Large \implies k=\huge \boxed 7$

$\dfrac{x-4}{1}=\dfrac{y-2}{1}=\dfrac{z-k}{2} \, (=\mu) \, \Rightarrow \, \textbf{r} = \begin{pmatrix} 4 \\ 2 \\ k \end{pmatrix} + \mu \begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix} \\ 2x-4y+z=7 \, \Rightarrow \, \textbf{r} \cdot \begin{pmatrix} 2 \\ -4 \\ 1 \end{pmatrix} = 7$ Substituting the line vector equation into the scalar product form of the plane, we get $8+2 \mu -8 -4 \mu +k+2 \mu = 7 \Rightarrow k=\large \color{#20A900}{\boxed{7}}$

$For\quad a\quad line\quad \frac { x-{ x }_{ 1 } }{ { a }_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { b }_{ 1 } } =\frac { z-{ z }_{ 1 } }{ { c }_{ 1 } } \\ Lies\quad in\quad the\quad plane\quad ax+by+cz=d\\ a{ .a }_{ 1 }+{ b.b }_{ 1 }+c.{ c }_{ 1 }=0\quad And\quad a{ .x }_{ 1 }+{ b.y }_{ 1 }+c.{ z }_{ 1 }=d\\ In\quad this\quad question\quad { a }_{ 1 }={ b }_{ 1 }=1\quad and\quad { c }_{ 1 }=2\\ { x }_{ 1 }{ =4,\quad y }_{ 1 }=2\quad ,\quad { z }_{ 1 }=k\quad a=2\quad ,\quad b=-4,\quad c=1,\quad d=7\\ now\quad 2.4+(-4).2+1.k=7\\ =>\quad k=7$