IF the straight lines $\frac{x-1}{k}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x-2}{3}=\frac{y-3}{k}=\frac{z-1}{2}$ intersect at a point, then integer k is equal to

-5

If given lines intersect, then they are coplanar.

$∴\begin{vmatrix}1-2 & 2-3 & 3-1\\k & 2 & 3\\3 & k & 2\end{vmatrix}=0$

$⇒ -(4-3k) + (2k - 9) + 2(k^2 - 6) = 0 $

$⇒ 2k^2 + 5k - 25 = 0 ⇒(k+5) (2k-5)= 0 ⇒ k = -5, 5/2.$