The vector equation of a plane passing through the line of intersection of the planes $\mathbf{r} \cdot(\hat{\mathbf{i}}-2 \hat{\mathbf{k}})=3, \mathbf{r} \cdot(2 \hat{\mathbf{j}}+\hat{\mathbf{k}})=5$ and the point $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ is

The points $A(-1,2,3), B(2,-3,1)$ and $C(3,1,-2)$

The directions cosines of the line making angles $\frac{\pi}{4}, \frac{\pi}{3}$ and $\theta\left(0<\theta<\frac{\pi}{2}\right)$ respectively with $X, Y$ and $Z$ axes are

If the equation of the plane passing through the point $(3,2,5)$ and perpendicular to the planes $2 x-3 y+5 z=7$ and $5 x+2 y-3 z=11$ is $x+b y+c z+d=0$, then $2 b+3 c+d=$