Find the equation of the plane passing through the origin and parallel to the plane $3x - 4y + 5z - 6 = 0$.

$A$ plane which bisects the angle between the two given planes $2x - y + 2z - 4 = 0$ and $x + 2y + 2z - 2 = 0$,passes through the point

The direction ratios of the two lines $AB$ and $AC$ are $1, -1, -1$ and $2, -1, 1$. The direction ratios of the normal to the plane $ABC$ are

Let $(\alpha, \beta, \gamma)$ be the image of the point $P (2, 3, 5)$ in the plane $2x + y - 3z = 6$. Then $\alpha + \beta + \gamma$ is equal to

The Cartesian equation of the plane passing through the point $(3, -2, -1)$ and parallel to the vectors $\vec{b} = \hat{i} - 2\hat{j} + 4\hat{k}$ and $\vec{c} = 3\hat{i} + 2\hat{j} - 5\hat{k}$ is:

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