If A and B are the feet of the perpendiculars drawn from $(1, 2, 3)$ to planes $YZ$ and $ZX$, then the equation of the plane passing through the points A, B and the origin is

If the distance of point $B(2,1,-3)$ from the line passing through the point $A(4,-2,2)$, and parallel to the vector $\vec{c} = -4\hat{i} - 6\hat{j} - 2\hat{k}$ is $x$, then $x^4 + x^2 + 541 =$

If $\theta$ is the angle between the lines $\dfrac{x-1}{2} = \dfrac{2y+3}{4};\ z = -2$ and $x = 1;\ \dfrac{y-1}{2} = \dfrac{z+1}{2}$, then

If for some $m \in \mathbb{R}$ the lines $L_1 : \dfrac{x+1}{m} = \dfrac{y-m}{-1} = \dfrac{z-1}{1}$ and $L_2 : \dfrac{x+2}{-4} = \dfrac{y+1}{9} = \dfrac{z+1}{1}$ are coplanar, then line $L_1$ passes through the point