Question

The plane passing through the point $(4,-1,2)$ and parallel to the lines $\frac{x+2}{3}=\frac{y-2}{-1}=\frac{z+1}{2}$ and $\frac{x-2}{1}=\frac{y-3}{2}=\frac{z-4}{3}$ also passes through the point.

• A. $(-1,-1,-1)$
• B. $(-1,-1,1)$
• C. $(1,1,-1)$
• D. $(1,1,1)$

Answer 1

The correct option is D $(1,1,1)$

Let $\overrightarrow{n}$ be the normal vector to the plane passing through $(4,-1,2)$ and parallel to the lines $L_1$ & $L_2$

then $\overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -1& 2\\ 1& 2& 3\end{array}\right|$

$\therefore \overrightarrow{n}=-7\hat{i}-7\hat{j}+7\hat{k}$
$\therefore$ Equation of plane is

$-1(x-4)-1(y+1)+1(z-2)=0$
$\therefore x+y-z-1=0$

Now check options.