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CBSE Class 12 Math 2026 All Sets Solved Paper

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Question : 14 of 20
Marks: +1, -0
The value of m for which the points with position vectors −i^−j^+2k^,2i^+mj^+5k^-\hat{i}-\hat{j}+2\hat{k}, 2\hat{i}+m\hat{j}+5\hat{k} and 3i^+11j^+6k^3\hat{i}+11\hat{j}+6\hat{k} are collinear, is
Solution:  
Let the three points be A, B & C
Their position vectors are
OA→=2i^+mj^+5k^\overrightarrow{OA}=2\hat{i}+m\hat{j}+5\hat{k}
OB→=3i^+11j^+6k^\overrightarrow{OB}=3\hat{i}+11\hat{j}+6\hat{k}
OC→=−i^−j^+2k^\overrightarrow{OC}=-\hat{i}-\hat{j}+2\hat{k}
AB→=OB→−OA→=(3−2)i^+(11−m)j^+(6−5)k^\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=(3-2)\hat{i}+(11-m)\hat{j}+(6-5)\hat{k}
=i^+(11−m)j^+k^=\hat{i}+(11-m)\hat{j}+\hat{k}
BC→=OC→−OB→=−4i^−12j^−4k^\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}=-4\hat{i}-12\hat{j}-4\hat{k}
Now,1−4=11−m−12=1−4Now, \frac{1}{-4}=\frac{11-m}{-12}=\frac{1}{-4}
⇒11−m−12=1−4\Rightarrow \frac{11-m}{-12}=\frac{1}{-4}
⇒11−m=3\Rightarrow 11-m=3
⇒m=11−3\Rightarrow m=11-3
⇒m=8\Rightarrow m=8
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