Determine the Constants and find the intersection points

Question

If $L_1$ is the line given by $at\hat i + (1+2t)\hat j + t\hat k$, $L_2$ is the line given by $(1 + t)\hat i + t\hat j + (1 + bt)\hat k$, and $S$ is the plane given by $x + 2y + z = 3$, find $a$ and $b$ and the intersection point of $S$ and $L_2$. $L_1$ and $L_2$ are parallel to each other. I honestly have no clue how to even start, any help would be appreciated.

Answer 1

Any point on $L_1$ will have position vector $$\hat j + t(a \hat i + 2\hat j + \hat k)$$ and any point on $L_2$ will have position vector $$\hat i + \hat k +t(\hat i + \hat j + b\hat k)$$

Since they are parallel, their direction vectors must be parallel (the vectors inside the bracket multiplied by $t$). Note that multiplying any vector by a scalar preserves its direction, and so we conclude that $L_1$ is proportional to $L_2$.

$$\implies \frac a1 = \frac 21 = \frac 1b$$ $$\implies a=2, \ b=\frac 12$$

Now, from the equation of $L_2$ we can deduce that the coordinates of any general point on $L_2$ are $$(1+t, t, 1+ \frac t2)$$. Plugging this point into our plane equation we get $$(1+t) + 2t + (1+\frac t2) = 3 \implies t=\frac 27$$ And so the intersection point is $(\frac 97, \frac 27, \frac 87)$.