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The plane passing through the point

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:

 

  1. $(-1,-1,-1)$

  2. $(-1,-1,1)$

  3. $(1,1,-1)$

  4. $(1,1,1)$


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

 

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