Let the line $L$ pass through the point $(0,1,2)$,intersect the line $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and be parallel to the plane $2x+y-3z=4$. Then the distance of the point $P(1,-9,2)$ from the line $L$ is

The direction ratios (d.r.s.) of the normal to the plane passing through the origin and the line of intersection of the planes $x+2y+3z=4$ and $4x+3y+2z=1$ are

If a line $L$ is the line of intersection of the planes $2x + 3y + z = 1$ and $x + 3y + 2z = 2$. If line $L$ makes an angle $\alpha$ with the positive $X$-axis,then the value of $\sec \alpha$ is

If the point of intersection of the lines $\frac{x+1}{3} = \frac{y+a}{5} = \frac{z+b+1}{7}$ and $\frac{x-2}{1} = \frac{y-b}{4} = \frac{z-2a}{7}$ lies on the $xy$-plane, then the value of $a+b$ is:

Let the foot of the perpendicular from the point $A(4, 3, 1)$ on the plane $P: x - y + 2z + 3 = 0$ be $N$. If $B(5, \alpha, \beta)$,where $\alpha, \beta \in \mathbb{Z}$,is a point on the plane $P$ such that the area of the triangle $ABN$ is $3\sqrt{2}$,then $\alpha^2 + \beta^2 + \alpha\beta$ is equal to $...........$.

The ratio in which the line joining $(2, -4, 3)$ and $(-4, 5, -6)$ is divided by the plane $3x + 2y + z - 4 = 0$ is