The equation of line parallel to the plane x+y+z=2 and intersecting the line x+y−z=0=x+2y−3z+5 at the point  -4,1,3 is

# The equation of line parallel to the plane $x+y+z=2$ and intersecting the line $x+y-z=0=x+2y-3z+5$ at the point  $\left(-4,1,3\right)$ is

1. A

$\frac{x+4}{-3}=\frac{y-1}{-2}=\frac{z-3}{1}$

2. B

$\frac{x+4}{1}=\frac{y-1}{2}=\frac{z-3}{-3}$

3. C

$\frac{x+4}{-3}=\frac{y-1}{2}=\frac{z-3}{1}$

4. D

$\frac{x+4}{-1}=\frac{y-1}{2}=\frac{z-3}{-3}$

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### Solution:

The equation of plane is $x+y-z+\lambda \left(x+2y-3z+5\right)=0$

It is passing through the point $\left(-4,1,3\right)$

It implies that$-4+1-3+\lambda \left(-4+2-9+5\right)=0⇒\lambda =-1$

Hence, the equation of plane is  $y-2z+5=0$

The required line is parallel to both planes $y-2z+5=0$ and $x+y+z=2$

Normal vectors to above planes are${n}_{1}=i+j+k,{n}_{2}=j-2k$

The vector along the line is perpendicular to both the vectors, so it is along  ${n}_{1}×{n}_{2}$

Hence,${n}_{1}×{n}_{2}=\left|\begin{array}{ccc}i& j& k\\ 1& 1& 1\\ 0& 1& -2\end{array}\right|=-3i+2j+k$

Hence, the direction ratios of the required line are $⟨-3,2,1⟩$

Therefore, the equation of the required line is  $\overline{)\frac{x+4}{-3}=\frac{y-1}{2}=\frac{z-3}{1}}$

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