The line passing through $(4, -1, 2)$ and $(-3, 2, 3)$ meets the plane at right angles at the point $(-10, 5, 4)$. Then the equation of the plane is:

The equation of a plane through the line of intersection of the planes $x + 2y = 3$ and $y - 2z + 1 = 0$,and perpendicular to the first plane $x + 2y = 3$ is:

$A$ plane $\pi$ passing through the points $2 \hat{i}-3 \hat{j}$ and $3 \hat{i}+4 \hat{k}$ is parallel to the vector $2 \hat{i}+3 \hat{j}-4 \hat{k}$. If a line joining the points $\hat{i}+2 \hat{j}$ and $\hat{j}-2 \hat{k}$ intersects the plane $\pi$ at the point $a \hat{i}+b \hat{j}+c \hat{k}$,then $a+b+2c=$

If the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies in the plane $x+3y-\alpha z+\beta=0$,then $(\alpha, \beta)=$

Let $P$ be a plane $lx + my + nz = 0$ containing the line $\frac{1-x}{1} = \frac{y+4}{2} = \frac{z+2}{3}$. If plane $P$ divides the line segment $AB$ joining points $A(-3, -6, 1)$ and $B(2, 4, -3)$ in the ratio $k : 1$,then the value of $k$ is equal to

If the angle $\theta$ between the line $\frac{x + 1}{1} = \frac{y - 1}{2} = \frac{z - 2}{2}$ and the plane $2x - y + \sqrt{\lambda} z + 4 = 0$ is such that $\sin \theta = \frac{1}{3}$,the value of $\lambda$ is