Solution
Step 1: Point P (Image) ke coordinates nikalna
Kisi point (x1,y1,z1) ki image (x,y,z) plane ax+by+cz+d=0 mein nikalne ka formula hota hai:
ax−x1=by−y1=cz−z1=−2a2+b2+c2(ax1+by1+cz1+d)
Yaha point (35,35,38) hai aur plane x−2y+z−2=0 hai.
Pehle RHS calculate karte hain:
−212+(−2)2+12(35−2(35)+38−2)=−26(35−10+8−6)=−26(−3/3)=−26(−1)=31
Ab x,y,z ke liye solve karte hain:
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1x−5/3=31⟹x=31+35=2
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−2y−5/3=31⟹y=−32+35=1
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1z−8/3=31⟹z=31+38=3
To point P=(2,1,3) mil gaya.
Step 2: Distance formula se α nikalna
Point P(2,1,3) aur Q(6,−2,α) ke beech ki doori 13 hai:
(6−2)2+(−2−1)2+(α−3)2=169
Isko solve karne par:
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α−3=12⟹α=15
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α−3=−12⟹α=−9
Chunki sawal mein diya hai ki \alpha > 0, isliye hum α=15 lenge.
Answer: 15
