MHTCET 25 Apr 2025 Shift 2 Paper

Concept:Two lines are coplanar if the scalar triple product of their direction vectors and the vector joining their points is zero.Explanation:Write each line in symmetric form.First line: $x = ay - 1 = z - 2$ gives $\frac{x}{1} = \frac{y - \frac{1}{a}}{\frac{1}{a}} = \frac{z - 2}{1}$.So point $P_1 = \left(0, \frac{1}{a}, 2\right)$ and direction $\vec{d_1} = \left(1, \frac{1}{a}, 1\right)$.Second line: $x = 3y - 2 = bz - 2$ gives $\frac{x}{1} = \frac{y - \frac{2}{3}}{\frac{1}{3}} = \frac{z - \frac{2}{b}}{\frac{1}{b}}$.So point $P_2 = \left(0, \frac{2}{3}, \frac{2}{b}\right)$ and direction $\vec{d_2} = \left(1, \frac{1}{3}, \frac{1}{b}\right)$.For coplanarity, $(\vec{P_2} - \vec{P_1}) \cdot (\vec{d_1} \times \vec{d_2}) = 0$.Hence determinant:$$\begin{vmatrix} 0 & \frac{2}{3} - \frac{1}{a} & \frac{2}{b} - 2 \\ 1 & \frac{1}{a} & 1 \\ 1 & \frac{1}{3} & \frac{1}{b} \end{vmatrix} = 0$$Expanding gives:$$-\left(\frac{2}{3} - \frac{1}{a}\right)\left(\frac{1}{b} - 1\right) + \left(\frac{2}{b} - 2\right)\left(\frac{1}{3} - \frac{1}{a}\right) = 0$$Simplifying:$$-\frac{1}{ab} + \frac{1}{a} = 0 \implies \frac{1}{a}\left(1 - \frac{1}{b}\right) = 0$$Since $ab \neq 0$, $a \neq 0$, so $1 - \frac{1}{b} = 0 \Rightarrow b = 1$.Also $a$ can be any non-zero real number.Answer:D. $b = 1, a \in \mathbb{R} - \{0\}$