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 A $(1,1,1)$

Normal vector of the plane is
$\overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -1& 2\\ 1& 2& 3\end{array}\right|$
$=\hat{i}(-3-4)-\hat{j}(9-2)+\hat{k}(6+1)$
$=7(-\hat{i}-\hat{j}+\hat{k})$
$\therefore$ the equation of plane $\left(\overrightarrow{r}\right)$ passing through the point $(4,-1,2)$ is
$(\overrightarrow{r}-\overrightarrow{a})\cdot \overrightarrow{n}=0$
$\Rightarrow (x-4)(-7)+(y+1)(-7)+(z-2)\left(7\right)=0$
$\Rightarrow x-4+y+1-z+2=0$
$\Rightarrow x+y-z-1=0$
Clearly, point $(1,1,1)$ lies on the plane