If $M$ is the foot of the perpendicular drawn from $P($ -1,2,-1 ) to the plane passing through the point $A(3,-2,1)$ and perpendicular to the vector $4 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-4 \hat{\mathbf{k}}$, then the length of $P M$ is
Vectors $\mathbf{p}=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}, \mathbf{q}=d \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$ and $\mathbf{r}=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ forming a $\triangle A B C$ are such that $\mathbf{p}=\mathbf{q}+\mathbf{r}$. If the area of $\triangle A B C$ is $5 \sqrt{6}$ sq. units, then the sum of the absolute values of $a, b, c$ is
$\mathbf{b}$ and $\mathbf{c}$ are non-collinear vectors and $(\mathbf{c} \cdot \mathbf{c}) \mathbf{a}=\mathbf{c}$. If $(\mathbf{a} \cdot \mathbf{c}) \mathbf{b}-(\mathbf{a} \cdot \mathbf{b}) \mathbf{c}+(\mathbf{a} \cdot \mathbf{b}) \mathbf{b}$ $=(4-2 \beta-\sin \alpha) \mathbf{b}+\left(\beta^2-1\right) \mathbf{c}$, then $\sin (\alpha+\beta)=$
Assertion (A) The variance of the first $n$ odd natural numbers is $\frac{n^2-1}{3}$.
Reason (R) The sum of the first $n$ odd natural numbers is $n^2$ and the sum of the squares of the first $n$ odd natural numbers is $\frac{n\left(4 n^2-1\right)}{3}$.
Which of the following alternatives is correct?