Let $d$ be the distance of the point of intersection of the lines $\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}$ and $\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}$ from the point $(7,8,9)$. Then $d^2+6$ is equal to :

Line $L$ passes through two points $(2, -3, 1)$ and $(3, -4, -5)$. If point $(0, a, b)$ lies on the line $L$,then $a+b =$ . . . . . . .

$A$ line passes through $A(4, -6, -2)$ and $B(16, -2, 4)$. The point $P(a, b, c)$,where $a, b, c$ are non-negative integers,lies on the line $AB$ at a distance of $21$ units from point $A$. The distance between the points $P(a, b, c)$ and $Q(4, -12, 3)$ is equal to...........

If the lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{-k}$ and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}$ are coplanar,then $k = . . . . .$

Let a line passing through the point $P(4,1,0)$ intersect the line $L_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ at the point $A(\alpha, \beta, \gamma)$ and the line $L_2: x-6=y=-z+4$ at the point $B(a, b, c)$. Then $\left|\begin{array}{lll}1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c \end{array}\right|$ is equal to

The coordinates of the point where the line joining the points $(3, 5, -7)$ and $(-2, 1, 8)$ is intersected by the $yz$-plane are given by: