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:
Correct Option: , 4
Solution:
Let $\overrightarrow{\mathrm{n}}$ be the normal vector to the plane passing through $(4,-1,2)$ and parallel to the lines $\mathrm{L}_{1} \& \mathrm{~L}_{2}$
then $\overrightarrow{\mathrm{n}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 3 & -1 & 2 \\ 1 & 2 & 3\end{array}\right|$
$\therefore \overrightarrow{\mathrm{n}}=-7 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}$
$\therefore$ Equation of plane is
$-1(x-4)-1(y+1)+1(z-2)=0$
$\therefore x+y-z-1=0$
Now check options