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3D Geometry question

2025 · 22 Jan · Shift 1 · Q48
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3D Geometry question

2025 · 22 Jan · Shift 1 · Q48

JEE MainMathematics3D GeometryNumerical+4 / −1
Let L1:x−13=y−1−1=z+10\mathrm{L}_1: \frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}L1​:3x−1​=−1y−1​=0z+1​ and L2:x−22=y0=z+4α,α∈R\mathrm{L}_2: \frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{\alpha}, \alpha \in \mathbf{R}L2​:2x−2​=0y​=αz+4​,α∈R, be two lines, which intersect at the point BBB. If PPP is the foot of perpendicular from the point A(1,1,−1)A(1,1,-1)A(1,1,−1) on L2L_2L2​, then the value of 26α( PB)226 \alpha(\mathrm{~PB})^226α( PB)2 is ‾\underline{\hspace{2cm}}​ .
Numerical answer
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