If the equation of the plane passing through the point $A(-2, 1, 3)$ and perpendicular to the vector $3 \hat{i} + \hat{j} + 5 \hat{k}$ is $ax + by + cz + d = 0$,then $\frac{a + b}{c + d} = $

$A$ point moves such that its distances from the points $(3, 4, -2)$ and $(2, 3, -3)$ are equal. What is the locus of the point?

The equation of the planes parallel to the plane $x - 2y + 2z - 3 = 0$ which are at a unit distance from the point $(1, 2, 3)$ is $ax + by + cz + d = 0$. If $(b - d) = K(c - a)$,then the positive value of $K$ is

If the planes $\bar{r} \cdot(2 \hat{i}-\lambda \hat{j}+\hat{k})=3$ and $\bar{r} \cdot(4 \hat{i}-\hat{j}+\mu \hat{k})=5$ are parallel,then $\lambda+\mu=$

Let $\pi_1$ be the plane passing through the point $2\hat{i}-\hat{j}+\hat{k}$ and perpendicular to the vector $a\hat{i}+2\hat{j}-3\hat{k}$, and $\pi_2$ be the plane passing through the point $\hat{i}+2\hat{j}-\hat{k}$ and perpendicular to the vector $\hat{i}-2\hat{j}+\hat{k}$. If $\theta$ is the angle between the planes $\pi_1$ and $\pi_2$ and $\cos \theta = -\sqrt{\frac{3}{7}}$, then the integral value of $a$ is:

Find the vector equation of a plane which is at a distance of $7$ units from the origin and normal to the vector $3 \hat{i} + 5 \hat{j} - 6 \hat{k}$.