Question

Let the equation of the plane through the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and parallel to the $x$-axis be $k(y-2)-m(z-3)=0$. Find $k+m$

Answer 1

Equation of a plane through the given line is
$A(x-1)+B(y-2)+C(z-3)=0$
where $2A+3B+4C=0.$
If the plane is parallel to $x$ - axis, i.e. $\perp$ to $yz$ - plane, i.e. $x=0;$
$A.1+B.0+C.0=0$

$\therefore A=0$
and hence $3B+4C=0$ or $\frac{B}{4}=\frac{C}{-3}.$
Thus the equation of the plane is $4(y-2)-3(z-3)=0$
$\Rightarrow 4y-3z+1=0.$
$\Rightarrow k+m=7$