The distance of the line $\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}$ from the point $(1,4,0)$ along the line $\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}$ is:

$A$ line $L$ is parallel to both the planes $2x + 3y + z = 1$ and $x + 3y + 2z = 2$. If line $L$ makes an angle $\alpha$ with the positive direction of the $X$-axis, then $\cos \alpha =$

The reflection of the point $(-1, 3, 4)$ with respect to the plane $x - 2y = 0$ is .....

Let $L_1$ be the line of intersection of the planes given by the equations $2x+3y+z=4$ and $x+2y+z=5$. Let $L_2$ be the line passing through the point $P(2,-1,3)$ and parallel to $L_1$. Let $M$ denote the plane given by the equation $2x+y-2z=6$. Suppose that the line $L_2$ meets the plane $M$ at the point $Q$. Let $R$ be the foot of the perpendicular drawn from $P$ to the plane $M$. Then which of the following statements is (are) True?
$(A)$ The length of the line segment $PQ$ is $9\sqrt{3}$
$(B)$ The length of the line segment $QR$ is $15$
$(C)$ The area of $\triangle PQR$ is $\frac{3}{2}\sqrt{234}$
$(D)$ The acute angle between the line segments $PQ$ and $PR$ is $\cos^{-1}\left(\frac{1}{2\sqrt{3}}\right)$

If a plane $P$ passes through the points $(1,0,0)$ and $(0,1,0)$ and makes an angle $\frac{\pi}{4}$ with the plane $x+y=3$, then the direction ratios of a normal to that plane $P$ are

Find the distance of the point whose position vector is $(2 \hat{i}+\hat{j}-\hat{k})$ from the plane $\vec{r} \cdot(\hat{i}-2 \hat{j}+4 \hat{k})=9$.