Question

Find the equation of line through $(-4,1,3)$ and parallel to the plane $x+y+z=3$ while the line intersects another line whose equation is $x+y+z=0=x+2y-3z+5$.

• A. $\frac{x+4}{-3}=\frac{y-1}{-2}=\frac{z-3}{1}$
• B. $\frac{x+4}{1}=\frac{y-1}{2}=\frac{z-3}{1}$
• C. $\frac{x+4}{-3}=\frac{y-1}{2}=\frac{z-3}{1}$
• D. $\frac{x+4}{-1}=\frac{y-1}{2}=\frac{z-3}{-3}$

Answer 1

The correct option is C

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

Simplify the given equation of the plane using the equation of the plane and the given point.

We know, the line of intersection of planes is

$$\begin{gathered} {\mathrm{\pi }}_1+{\mathrm{\lambda \pi }}_2=0 \\ \Rightarrow (x+y-z)+\lambda (x+2y-3z+5)=0 \end{gathered}$$

Here, it passes through $(-4,1,3)$

Therefore, $\lambda =-1$.

Clearly, the equation of the plane will be $y-2z+5=0$.

As the required line lies in this plane and is parallel to $x+y+z=3$,

therefore $$\begin{gathered} =\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ 0& 1& -2\end{array}\right| \\ =-3\hat{i}+2\hat{j}+\hat{k} \end{gathered}$$

Hence, required line will be $=\frac{x+4}{-3}=\frac{y-1}{2}=\frac{z-3}{1}$

Therefore, Option(C) is the correct answer.