$A$ plane $\Pi$ passes through the points $A=(0,0,2)$, $B=(1,0,1)$, and $C=(3,1,1)$. If the plane $\Pi$ makes angles $\alpha$ and $\beta$ with the $XY$ and $XZ$-coordinate planes respectively, then $\sin^2 \alpha + \sin^2 \beta =$

Find the angle between the two planes $2x + y - 2z = 5$ and $3x - 6y - 2z = 7$ using the vector method.

If the vector equation of the plane $\bar{r}=(2 \hat{i}+\hat{k})+\lambda \hat{i}+\mu(\hat{i}+2 \hat{j}-3 \hat{k})$ in scalar product form is given by $\bar{r} \cdot(3 \hat{j}+2 \hat{k})=\alpha$,then $\alpha=$

If $(x, y, z)$ is an arbitrary point lying on a plane $P$ which passes through the points $(42, 0, 0)$,$(0, 42, 0)$,and $(0, 0, 42)$,then the value of the expression $3 + \frac{x-11}{(y-19)^{2}(z-12)^{2}} + \frac{y-19}{(x-11)^{2}(z-12)^{2}} + \frac{z-12}{(x-11)^{2}(y-19)^{2}} - \frac{x+y+z}{14(x-11)(y-19)(z-12)}$ is:

$(1, -2, 1)$ is a point on a plane $\pi$ and $\pi$ is parallel to the plane $x-y-z=0$. If the equation of $\pi$ is $ax+by+cz-2=0$, then $b-2c=$

If the foot of the perpendicular from $(0,0,0)$ to a plane is $(1,2,3)$, then the equation of the plane is