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If two lines x−32=y−45=z4\frac{x-3}{2} = \frac{y-4}{5} = \frac{z}{4} and x−43=y−56=1−zk\frac{x-4}{3} = \frac{y-5}{6} = \frac{1-z}{k}, are coplanar, then kk is equal to

Solution

✅ Correct Option: 2

Direction of L2L_2: (3,6,−k)(3, 6, -k) since 1−zk=z−1−k\frac{1-z}{k} = \frac{z-1}{-k}.

Vector joining points: (1,1,1)(1, 1, 1).

Coplanarity: scalar triple product = 0.

∣11125436−k∣=−3k−15=0\begin{vmatrix} 1 & 1 & 1 \\ 2 & 5 & 4 \\ 3 & 6 & -k \end{vmatrix} = -3k - 15 = 0, so k=−5k = -5.

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