QJEEM23Ma3bb6: JEE Main 2023 Mathematics - Class 12 JEEM23Ma3bb6 Solution: If the lines and intersect, then the magnitude of the minimum…

Given, lines

$\begin{aligned}& \frac{x-1}{2} =\frac{2-y}{-3}=\frac{z-3}{\alpha} \\\\&\Rightarrow \frac{x-1}{2} =\frac{y-2}{3}=\frac{z-3}{\alpha}=\lambda .........(i)\end{aligned}$

Any point on the line (i)

$x=2 \lambda+1, y=3 \lambda+2, z=\alpha \lambda+3$

and line $\frac{x-4}{5}=\frac{y-1}{2}=\frac{z}{\beta}=\mu$ ............(ii)

Any point on line (ii)

$\Rightarrow x=5 \mu+4, y=2 \mu+1, z=\beta \mu$

Since, given lines intersects

$\begin{aligned}& \therefore 2 \lambda+1=5 \mu+4 ..........(iii)\\\\& 3 \lambda+2=2 \mu+1 ............(iv)\\\\& \text { and } \alpha \lambda+3=\beta \mu ..........(iv)\end{aligned}$

On solving (iii) and (iv), we get

$\lambda=-1, \mu=-1$

On putting value of $\lambda$ and $\mu$ in (v), we get

$\begin{array}{cc} & \alpha(-1)+3=-\beta \\\\&\Rightarrow \alpha=\beta+3\end{array}$

Now,

$\begin{aligned}& 8 \alpha \beta=8 (\beta+3)(\beta) \\\\&= 8\left(\beta^2+3 \beta\right) \\\\& =8\left(\beta^2+3 \beta+\frac{9}{4}-\frac{9}{4}\right) \\\\& =8\left\{\left(\beta+\frac{3}{2}\right)^2-\frac{9}{4}\right\}\end{aligned}$

$=8\left(\beta+\frac{3}{2}\right)^2-18$

Here, minimum value $=-18$

$\therefore$ Magnitude of the minimum value of $8 \alpha \beta$ is 18 .