The lines x-2/1=y-3/1=z-4/-k and x-1/k=y-4/2=z-5/1 are coplanar if

Correct Answer is : k=0,−3k=0,-3k=0,−3

Correct Answer is : k=0,−3k=0,-3k=0,−3

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MHTCET 16 Apr 2026 Shift 1 Paper

Question : 140 of 150

Marks: +1, -0

The lines

\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}

\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}

are coplanar if

Solution:

Concept:

Two lines are coplanar when the scalar triple product of their direction vectors and the vector joining their points is zero.

Explanation:

Line 1 passes through

A(2,3,4)

with direction vector

\vec{d_1}=(1,1,-k)

Line 2 passes through

B(1,4,5)

with direction vector

\vec{d_2}=(k,2,1)

Vector

\overrightarrow{AB}=B-A=(-1,1,1)

For coplanarity,

\overrightarrow{AB} \cdot (\vec{d_1} \times \vec{d_2})=0

This gives the determinant:

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

Expanding:

-(1+2k) - (1+k^2) + (2-k) = 0

Simplifying:

-k^2 - 3k = 0

-k(k+3) = 0

Thus

k=0

k=-3

Answer:

k=0,-3

So the correct option is A.

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