The vector equation of the plane passing through the point $A(-2, 7, 5)$ and parallel to the vectors $\vec{b} = 4\hat{i} - \hat{j} + 3\hat{k}$ and $\vec{c} = \hat{i} + \hat{j} + \hat{k}$ is:

Let the plane passing through the point $(2,1,-1)$ and containing the line joining the points $(1,3,2)$ and $(1,2,1)$ make intercepts $p, q, r$ on the coordinate axes. Then $p+q+r=$

Find the vector and Cartesian equations of the plane that passes through the point $(1, 4, 6)$ and the normal vector to the plane is $\hat{i} - 2\hat{j} + \hat{k}$.

If the plane $-4x - 2y + 2z + \alpha = 0$ is at a distance of $2$ units from the plane $2x + y - z + 1 = 0$, then the product of all the possible values of $\alpha$ is

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:

If a plane cuts off intercepts $OA = a, OB = b, OC = c$ from the coordinate axes,then the area of the triangle $ABC$ is: