JEE MAIN - Mathematics (2025 - 3rd April Morning Shift - No. 10)
Let a line passing through the point $(4,1,0)$ intersect the line $\mathrm{L}_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ at the point $A(\alpha, \beta, \gamma)$ and the line $\mathrm{L}_2: x-6=y=-z+4$ at the point $B(a, b, c)$. Then $\left|\begin{array}{lll}1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c\end{array}\right|$ is equal to
16
6
8
12
Explanation
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$$\begin{aligned} & A=(\alpha, \beta, 4)=(2 \lambda+1,3 \lambda+2,4 \lambda+3) \\ & \text { and } B=(a, b, c)=(\mu+6, \mu,-\mu+4) \\ & \because \frac{\mu+2}{2 \lambda-3}=\frac{\mu-1}{3 \lambda+1}=\frac{-\mu+4}{4 \lambda+3} \\ & \therefore \quad \lambda \mu+8 \lambda+4 \mu=1 \quad \ldots(1)\\ & \text { and } 7 \lambda \mu-16 \lambda+4 \mu=7 \quad \ldots(2) \end{aligned}$$
from equation (1) and (2)
$$\lambda=-1 \text { and } \mu=3$$
Or $\lambda=-\frac{1}{3}$ and $\mu=1$
By taking, $\lambda=-1$ and $\mu=3$, we get
$$\begin{aligned} & \therefore \quad A(\alpha, \beta, 4)=(-1,-1,-1) \\ & \text { and } B(a, b, c)=(9,3,1) \end{aligned}$$
$$\therefore\left|\begin{array}{ccc} 1 & 0 & 1 \\ -1 & -1 & -1 \\ 9 & 3 & 1 \end{array}\right|=8$$
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