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Question : 34 of 160
Marks:
+1,
-0
Solution:
Let,
A(2a+3b−c).B(a−2b+3c) C(3a+4b−2c) and
D(ka−6b+6c) ∴=−a−5b+4c =−a+b−c =(k−2)a−9b+7c It is given
A,B,C,D are coplanar.
|| −1 | −5 | 4 |
| 1 | 1 | −1 |
| k−2 | −9 | 7 |
|=0 (k−2)(5−4)+9(1−4) +7(−1+5)=0 k−1=0 k=1
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