$\frac{a-1}{2}=\frac{b-3}{-1}=\frac{c-4}{1}=\lambda $$\Rightarrow a=2 \lambda+1 $$b=3-\lambda $$c=4+\lambda $$P \equiv\left(\lambda+1,3-\frac{\lambda}{2}, 4+\frac{\lambda}{2}\right) $$2(\lambda+1)-\left(3-\frac{\lambda}{2}\right)+\left(4+\frac{\lambda}{2}\right)+3=0$$2 \lambda+2-3+\frac{\lambda}{2}+4+\frac{\lambda}{2}+3=0$$3 \lambda+6=0$$ \Rightarrow\, \lambda=-2$$a=-3, \,b=5, \,c=2$
So the equation of the required line is
$\frac{x+3}{3}=\frac{y-5}{1}=\frac{z-2}{-5}$
In a plane, the equation of a line is given by the popular equation y = m x + C. Let's look at how the equation of a line is written in vector form and Cartesian form.
Vector Equation
Consider a line that passes through a given point, say ‘A’, and the line is parallel to a given vector '\(\vec{b}\)‘. Here, the line ’l' is given to pass through ‘A’, whose position vector is given by '\(\vec{a}\)‘. Now, consider another arbitrary point ’P' on the given line, where the position vector of 'P' is given by '\(\vec{r}\)'.
\(\vec{AP}\)=𝜆\(\vec{b}\)
Also, we can write vector AP in the following manner:
