The plane containing these two lines $\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$ and $\frac{x-2}{1}=\frac{y-4}{4}=\frac{z-6}{7}$, is
Answer & explanation
Correct answer: option 1
The equation of the plane containing the given lines is
$\begin{vmatrix}x+1 & y+3 & z+5\\3 & 5 & 7\\1 & 4 & 7\end{vmatrix}=0 ⇒ x- 2y + z = 0 $