The line $(x,y,z)=(2,-3,4)+\lambda(1,2,-3)$ intersects $2x+3y-z=13$ at $P$ and $yz$-plane at $Q$. If $PQ=a\sqrt{b}$, $a,b\in\mathbb{N}$, $a>3$, then $\dfrac{a+b}{3}$ equals
Let $\vec{a}$, $\vec{b}$, $\vec{c}$ have magnitudes 1, 2, 3 respectively, satisfying $|(\vec{a}\times\vec{b})\cdot\vec{c}|=6$. If $\vec{d}$ is a unit vector coplanar with $\vec{b}$ and $\vec{c}$ such that $\vec{b}\cdot\vec{d}=1$, then $|(\vec{a}\times\vec{c})\cdot\vec{d}|^2+|(\vec{a}\times\vec{c})\times\vec{d}|^2$ is
The general coordinates of a point R which divides the line joining A(3, –2, 4) and B(1, 1, 1) in the ratio μ : 1 are \(\left(\dfrac{\mu+3}{\mu+1},\ \dfrac{\mu-2}{\mu+1},\ \dfrac{\mu+4}{\mu+1}\right)\). If C(–1, 4, –2) lies on the line AB, then the value of |μ| is:
Let $OA$, $OB$, $OC$ be coterminous edges of a cuboid. If $l,m,n$ be the shortest distance between the sides $OA$, $OB$, $OC$ and their respective skew body diagonals to them, then find $\dfrac{\left(\frac{1}{l^2}+\frac{1}{m^2}+\frac{1}{n^2}\right)}{\left(\frac{1}{OA^2}+\frac{1}{OB^2}+\frac{1}{OC^2}\right)}$
Let $\vec{a}$, $\vec{b}$, $\vec{c}$ have magnitudes 1, 2, 3 respectively, satisfying $|(\vec{a}\times\vec{b})\cdot\vec{c}|=6$. If $\vec{d}$ is a unit vector coplanar with $\vec{b}$ and $\vec{c}$ such that $\vec{b}\cdot\vec{d}=1$, then $|(\vec{a}\times\vec{c})\cdot\vec{d}|^2+|(\vec{a}\times\vec{c})\times\vec{d}|^2$ is
Three mutually perpendicular lines have direction ratios $(1,1,0)$, $(1,0,1)$, $(0,1,1)$ and pass through $(1,0,1)$, $(1,1,0)$, $(0,1,1)$ respectively. The volume of the tetrahedron formed by them and the plane $x+y+z=1$ is $V$. Find $16V^2$.