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 foot of the perpendicular drawn from a point $A(1,1,1)$ onto a plane $\pi$ is $P(-3,3,5)$. If the equation of the plane parallel to the plane $\pi$ and passing through the midpoint of $AP$ is $ax-y+cz+d=0$,then $a+c-d=$

The plane passing through the line $L: \ell x-y+3(1-\ell)z=1, x+2y-z=2$ and perpendicular to the plane $3x+2y+z=6$ is $3x-8y+7z=4$. If $\theta$ is the acute angle between the line $L$ and the $y$-axis,then $415 \cos^{2} \theta$ is equal to...

The value of $k$,such that the line $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2}$ lies on the plane $2x-4y+z=7$,is

Find the vector equation of the plane passing through the intersection of the planes $\vec{r} \cdot (3\hat{i} - \hat{j} + \hat{k}) = 1$ and $\vec{r} \cdot (\hat{i} + 4\hat{j} - 2\hat{k}) = 2$,and passing through the point $\hat{i} + 2\hat{j} - \hat{k}$.

If the distance of the point $P(1, -2, 1)$ from the plane $x + 2y - 2z = \alpha$,where $\alpha > 0$,is $5$ units,then the foot of the perpendicular from $P$ to the plane is: