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The lines x-22= y-2= z-716 and x+34= y+23= z+21 intersect at the point P. If the distance of P from the line x+12= y-13= z-11 is l, then 14l2 is equal to .
108
Line 1: (2+2 , -2 , 7+16 )
Line 2: (-3+4 , -2+3 , -2+ )
x: 2+2 =-3+4 ; y: -2 =-2+3 =1, =- 12
(z: 7-8=-1=-2+1 ✓) P =(1,1,-1)
[figure]
AP=2 i-2 k, AP2=8
Projection AM= AP d| d|= 4+0-2 14= 2 14
l2=AP2-AM2=8- 414= 10814
14l2=108
Answer: 108
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