If a plane cuts the coordinate axes at $A, B$ and $C$ respectively such that the centroid of the triangle $ABC$ is $(6, 6, 3)$,then find the equation of that plane.

The equation of the plane passing through a point $A(2, -1, 3)$ and parallel to the vectors $\vec{a} = (3, 0, -1)$ and $\vec{b} = (-3, 2, 2)$ is:

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:

Find the vector and Cartesian equations of the plane that passes through the point $(1, 0, -2)$ and the normal to the plane is $\hat{i} + \hat{j} - \hat{k}$.

The equation of the plane which is parallel to the $xy$-plane and cuts an intercept of length $3$ from the $z$-axis is

The equation of the plane passing through $(1,0,0)$ and $(0,1,0)$ and making an angle of $45^{\circ}$ with the plane $x+y-3=0$ is: