The equation of the plane passing through $(4,4,0)$ and perpendicular to the planes $2x+y+2z+3=0$ and $3x+3y+2z-8=0$ is

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 $\alpha=$

From a point $P(a, b, c)$, perpendiculars $PA$ and $PB$ are drawn to $XY$ plane and $ZX$ plane respectively. If $O$ is the origin, then the equation of plane $OAB$ is

The direction ratios of the normal to the plane passing through the points $(1, 0, 0)$ and $(0, 1, 0)$ and making an angle of $\pi/4$ with the plane $x + y = 3$ are proportional to:

The mirror image of the point $P(-1, 2, -4)$ in the plane $x - y - 2z + 1 = 0$ is

Direction ratios of the normal to the plane passing through the points $A(0, 1, 1)$,$B(1, 1, 2)$,and $C(-1, 2, -2)$ are