# The equation of the plane passing through the point $(3,2,0)$ and the line $\large\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}$ is ?

$\begin{array}{1 1}(a)\:\:x-y+z=1\:\:\:\qquad\:\:(b)\:\:x+y+z=5\:\:\:\qquad\:\:(c)\:\:x+2y-z=1\:\:\:\qquad\:\:(d)\:\:2x-y+z=5. \end{array}$

Let the equation of the plane be $ax+by+cz+d=0$
$\Rightarrow\:$ The normal to the plane $\overrightarrow n$ is $(a,b,c)$
Since the plane passes through $(3,2,0)$, $3a+2b+d=0$.....(i)
Also the plane passes through the line $\large\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}$
$\Rightarrow\:3a+6b+4c+d=0$......(ii) and
since the line lies on the plane, the line is $\perp$ to the normal $\overrightarrow n$
$\therefore (a,b,c).(1,5,4)=0$
$\Rightarrow\:a+5b+4c=0$..........(iii)
$(ii)-(i)\:\:\Rightarrow\:4b+4c=0$.....(iv)
Solving $(iii)\:\:and\:\:(iv)$ using cross product method,
$\large\frac{a}{20-16}=\frac{b}{0-4}=\frac{c}{4-0}$
$\Rightarrow\:(a,b,c)=(4,-4,4)=(1,-1,1)$
Substituting the values of $a,b,c$ in (i) we get $d=-1$
$\therefore$ The equation of the plane is $x-y+z=1$