A line passes through M_1(2,4,λ) and the point (-3,0,0) of…

A line passes through $M_1(2,4,\lambda)$ and the point $(-3,0,0)$ of the plane $x+2y+13z+3=0$. Write its equation and determine $\lambda$ so that the line lies entirely in the plane.

Step 1: The line

The point $(-3,0,0)$ lies in the plane: $-3+3=0$. The direction is $\vec d=\{5,4,\lambda\}$: $\frac{x+3}{5}=\frac y4=\frac z\lambda$.

For $\lambda=0$ read the last fraction as $z=0$.

Step 2: Line in the plane

Since one point already lies in the plane, it suffices that $\vec d\perp\vec n$: $5+8+13\lambda=0\ \Longrightarrow\ \lambda=-1$.

\[\lambda=-1:\qquad \frac{x+3}{5}=\frac y4=\frac z{-1}\]

Check: $M_1(2,4,-1)$ gives $2+8-13+3=0$. Equivalently, one can require $M_1$ to lie in the plane: $2+8+13\lambda+3=0$, which gives the same $\lambda=-1$.