Question

If $ax+by+cz-1=0$ is the plane passing through $(4,-1,2)$ and parallel to the lines $\frac{x+2}{3}=\frac{y-2}{-1}=\frac{z+1}{2}$ and $\frac{x-2}{1}=\frac{y-3}{2}=\frac{z-4}{3},$ then $4a+3b+2c=$

Answer 1

Equation of plane passing through $(4,-1,2)$ is: $a(x-4)+b(y+1)+c(z-2)=0$
and parallel to the lines,
$\frac{x+2}{3}=\frac{y-2}{-1}=\frac{z+1}{2}$ and $\frac{x-2}{1}=\frac{y-3}{2}=\frac{z-4}{3},$
$\Rightarrow D.r^{\prime }s$ of $L_1=(3,-1,2)$
and $D.r^{\prime }s$ of $L_2=(1,2,3)$
So, $D.r^{\prime }s$ of normal to the plane is :
$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -1& 2\\ 1& 2& 3\end{array}\right|=-7\hat{i}-7\hat{j}+7\hat{k}$
$D.r^{\prime }s$ of normal $=(-7,-7,7)\equiv (-1,-1,1)$
So, equation of plane is $-(x-4)-(y+1)+(z-2)=0$
$\Rightarrow x+y-z-1=0$
$\Rightarrow a=1,b=1,c=-1$
$\therefore 4a+3b+2c=5$