QJEEM25Mad259: JEE Main 2025 Mathematics - Class 12 JEEM25Mad259 Solution: Let and be two lines. Let be a line passing through the point…

Step 1. Identify the Given Lines

The line

$\mathrm{L}_1:\; \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2}$

passes through the point

$P_1=(1,2,1)$

with direction vector

$\mathbf{u}=(1,-1,2).$

Similarly, the line

$\mathrm{L}_2:\; \frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1}$

passes through the point

$P_2=(-1,2,0)$

with direction vector

$\mathbf{v}=(-1,2,1).$

Step 2. Determine the Direction of $L_3$

Since $L_3$ is perpendicular to both $L_1$ and $L_2$ , its direction vector must be parallel to the cross product of $\mathbf{u}$ and $\mathbf{v}$ . Compute:

$\mathbf{d} = \mathbf{u}\times\mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & 2 \\ -1 & 2 & 1 \end{vmatrix}.$

Using the determinant, we obtain:

$\begin{aligned} d_x &= (-1)(1) - (2)(2) = -1-4=-5,\\[1mm] d_y &= (2)(-1) - (1)(1) = -2-1=-3,\\[1mm] d_z &= (1)(2) - (-1)(-1) = 2-1=1. \end{aligned}$

Thus, a valid direction vector is

$\mathbf{d}=(-5,-3,1).$

Step 3. Express $L_3$ in Terms of Its Intersection with $L_1$

Since $L_3$ intersects $L_1$ , let the intersection point (on $L_1$ ) be expressed using a parameter $t$ as:

$A(t) = (1+t,\, 2-t,\, 1+2t).$

Since $L_3$ passes through $A(t)$ and also through the given point

$Q=(\alpha,\beta,\gamma),$

its parametric form can be written as:

$(\alpha,\beta,\gamma)=A(t) + s\,\mathbf{d} = (1+t,\,2-t,\,1+2t) + s(-5,-3,1)$

for some parameters $s$ and $t$ .

Step 4. Express $\alpha,\beta,\gamma$ in Terms of $t$ and $s$

Comparing coordinates, we have:

$\begin{cases} \alpha = 1 + t - 5s,\\[1mm] \beta = 2 - t - 3s,\\[1mm] \gamma = 1 + 2t + s. \end{cases}$

Step 5. Compute $5\alpha - 11\beta - 8\gamma$

Substitute the expressions for $\alpha$ , $\beta$ , and $\gamma$ :

$\begin{aligned} 5\alpha - 11\beta - 8\gamma &= 5(1+t-5s) - 11(2-t-3s) - 8(1+2t+s)\\[1mm] &= (5 + 5t - 25s) - (22 - 11t - 33s) - (8 + 16t + 8s)\\[1mm] &= \left(5 - 22 - 8\right) + \left(5t + 11t - 16t\right) + \left(-25s + 33s - 8s\right)\\[1mm] &= -25 + 0t + 0s\\[1mm] &= -25. \end{aligned}$

Taking the absolute value yields:

$\left|5\alpha-11\beta-8\gamma\right| = 25.$

Conclusion

The value of

$\left|5\,\alpha - 11\,\beta - 8\,\gamma\right|$ is $\boxed{25}.$