If $$\mathbf{a}$$ is collinear with $$\mathbf{b}=3 \hat{i}+6 \hat{j}+6 \hat{k}$$ and $$\mathbf{a} \cdot \mathbf{b}=27$$, then $$|\mathbf{a}|=$$
Let $$a, b$$ and $$c$$ be unit vectors such that $$a$$ is perpendicular to the plane containing $$\mathbf{b}$$ and $$\mathbf{c}$$ and angle between $$\mathbf{b}$$ and $$\mathbf{c}$$ is $$\frac{\pi}{3}$$. Then, $$|\mathbf{a}+\mathbf{b}+\mathbf{c}|=$$
Let $$\mathbf{F}=2 \hat{i}+2 \hat{j}+5 \hat{k}, A=(1,2,5), B=(-1,-2,-3)$$ and $$\mathbf{B A} \times \mathbf{F}=4 \hat{i}+6 \hat{j}+2 \lambda \hat{k}$$, then $$\lambda=$$
If the mean deviation of the data $$1,1+d 1+2 d, \ldots, 1+100 d,(d>0)$$ from their mean is 255, then '$$d$$' is equal to
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