The vertices B and C of a triangle ABC lie on the line $\frac{x}{1}=\frac{1-y}{-2}=\frac{\mathrm{z}-2}{3}$. The coordinates of A and $B$ are $(1,6,3)$ and $(4,9, \alpha)$ respectively and $C$ is at a distance of 10 units from $B$. The area (in sq. units) of $\triangle A B C$ is :

Let the mean and variance of 8 numbers $-10,-7,-1, x, y, 9,2,16$ be $\frac{7}{2}$ and $\frac{293}{4}$, respectively.

Then the mean of 4 numbers $x, y, x+y+1,|x-y|$ is :

Number of solutions of $\sqrt{3} \cos 2 \theta+8 \cos \theta+3 \sqrt{3}=0, \theta \in[-3 \pi, 2 \pi]$ is :

Let $\alpha$ and $\beta$ respectively be the maximum and the minimum values of the function $f(\theta)=4\left(\sin ^4\left(\frac{7 \pi}{2}-\theta\right)+\sin ^4(11 \pi+\theta)\right)-2\left(\sin ^6\left(\frac{3 \pi}{2}-\theta\right)+\sin ^6(9 \pi-\theta)\right), \theta \in \mathbf{R}$.

Then $\alpha+2 \beta$ is equal to :