Find the equation of the plane with intercept $3$ on the $y$-axis and parallel to the $ZOX$ plane.

The distance between two parallel planes $ax+by+cz+d_1=0$ and $ax+by+cz+d_2=0$ is given by $\frac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}}$. If the plane $2x-y+2z+3=0$ is at distances of $\frac{1}{3}$ and $\frac{2}{3}$ units from the planes $4x-2y+4z+\lambda=0$ and $2x-y+2z+\mu=0$ respectively,then the maximum value of $\lambda+\mu$ is:

If the cartesian equation of the plane passing through the point $\hat{i}+2 \hat{j}+\hat{k}$ and parallel to the vectors $2 \hat{i}+3 \hat{j}+\hat{k}$ and $-\hat{i}+2 \hat{j}-3 \hat{k}$ is $a x+b y+c z=1$, then $18(a+b+c)$ is equal to

If the image of the point $A(1, 1, 1)$ with respect to the plane $4x + 2y + 4z + 1 = 0$ is $B(\alpha, \beta, \gamma)$,then $\alpha + \beta + \gamma =$

The mirror image of the point $(1, 2, 3)$ in a plane is $\left(-\frac{7}{3}, -\frac{4}{3}, -\frac{1}{3}\right)$. Which of the following points lies on this plane?

The equation of the plane passing through the point $(2, -1, -3)$ and parallel to the lines $\frac{x - 1}{3} = \frac{y + 2}{2} = \frac{z}{-4}$ and $\frac{x}{2} = \frac{y - 1}{-3} = \frac{z - 2}{2}$ is