Let the vectors $$\vec{u}_{1}=\hat{i}+\hat{j}+a \hat{k}, \vec{u}_{2}=\hat{i}+b \hat{j}+\hat{k}$$ and $$\vec{u}_{3}=c \hat{i}+\hat{j}+\hat{k}$$ be coplanar. If the vectors $$\vec{v}_{1}=(a+b) \hat{i}+c \hat{j}+c \hat{k}, \vec{v}_{2}=a \hat{i}+(b+c) \hat{j}+a \hat{k}$$ and $$\vec{v}_{3}=b \hat{i}+b \hat{j}+(c+a) \hat{k}$$ are also coplanar, then $$6(\mathrm{a}+\mathrm{b}+\mathrm{c})$$ is equal to :
If the probability that the random variable $$\mathrm{X}$$ takes values $$x$$ is given by $$\mathrm{P}(\mathrm{X}=x)=\mathrm{k}(x+1) 3^{-x}, x=0,1,2,3, \ldots$$, where $$\mathrm{k}$$ is a constant, then $$\mathrm{P}(\mathrm{X} \geq 2)$$ is equal to :
Let $$\mathrm{A}=\{1,2,3,4,5,6,7\}$$. Then the relation $$\mathrm{R}=\{(x, y) \in \mathrm{A} \times \mathrm{A}: x+y=7\}$$ is :
Let the mean and variance of 12 observations be $$\frac{9}{2}$$ and 4 respectively. Later on, it was observed that two observations were considered as 9 and 10 instead of 7 and 14 respectively. If the correct variance is $$\frac{m}{n}$$, where $$\mathrm{m}$$ and $$\mathrm{n}$$ are coprime, then $$\mathrm{m}+\mathrm{n}$$ is equal to :