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(B) Let the direction ratios of the three lines be $\vec{a} = (1, -1, 1)$, $\vec{b} = (2, -3, 0)$, and $\vec{c} = (1, 0, 3)$.Since all three lines pas

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coplanar

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(B) Let the direction ratios of the three lines be $\vec{a} = (1, -1, 1)$, $\vec{b} = (2, -3, 0)$, and $\vec{c} = (1, 0, 3)$.Since all three lines pass through the origin, they are coplanar if and only if the scalar triple product of their direction vectors is zero, i.e.,$[\vec{a}, \vec{b}, \vec{c}] = 0$.We calculate the determinant of the matrix formed by these vectors:$\Delta = \begin{vmatrix} 1 & -1 & 1 \ 2 & -3 & 0 \ 1 & 0 & 3 \end{vmatrix}$$= 1((-3)(3) - (0)(0)) - (-1)((2)(3) - (0)(1)) + 1((2)(0) - (-3)(1))$$= 1(-9 - 0) + 1(6 - 0) + 1(0 + 3)$$= -9 + 6 + 3 = 0$Since the determinant is $0$, the three vectors are linearly dependent, which means the three lines lie in the same plane.Therefore, the lines are coplanar.

Three lines are drawn from the origin $O$ with direction ratios proportional to $(1, -1, 1)$, $(2, -3, 0)$, and $(1, 0, 3)$. The three lines are

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