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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 |
$x - 2y + z= 0$ $x + 2y - z= 0$ $x - 2y + z= 1$ none of these |
$x - 2y + z= 0$ |
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 $ |
