If the distance between points ( α , 5 α , 10 α ) from the point of intersection of the line.
= (2 
− 
+ 2 
) + λ (2 
+ 4 
+ 12 
) and plane 
. ( 
− 
+ 
) = 5 is 13 units, then value of α may be
Text Solution
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(b,d)
Line : 
= 
⇒ 
= r (Let)
⇒ x = r + 2 ⇒ y = 2r – 1 ⇒ z = 6r + 2
plane : x – y + z = 5 ⇒ r + 2 – (2r – 1) + 6r + 2 = 5 ⇒ 5r = 0 ⇒ r = 0
∴ point of intersection is (2, – 1, 2) ⇒ ( α – 2) 2 + (5 α + 1) 2 + (10 α – 2) 2 = 169
126 α 2 – 34 α – 160 = 0 ⇒ 63 α 2 – 17 α – 80 = 0 ⇒ α = – 1, 
.
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