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Question : 31 of 160
Marks:
+1,
-0
Solution:
Let
1,m,n be
,, and
C(+4−3) Now,
A,B,C lie on line
L ∴‌‌AB=(2−p)−2+2 BC=−−+ A,B,C are collinear
∴‌‌‌=‌=‌ 2−p=−2⇒p=4 d.r. of line
(−1,−1,1) Equation of plane through part
(−p,p,p+1) i.e.,
(−4,4,5) || x−2 | y−5 | z+4 |
| −6 | −1 | 9 |
| −1 | −1 | 1 |
|=0 (x−2)(−1+9)−(y−5)(−6+9)+(z+4)(6−1)=0
8x−3y+5z=−19 −‌x+‌y−‌z=1 ∴‌‌a=−‌,b=‌,c=‌  
∵p(a+b+c)
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