If the straight lines $\vec{r} = (1, 2, 3) + k(\lambda, 2, 3), k \in R$ and $\vec{r} = (2, 3, 1) + k(3, \lambda, 2), k \in R$ intersect at a point,then the integer $\lambda$ is equal to:

The angle between the lines $x=y, z=0$ and $y=0, z=0$ is (in $^{\circ}$)

Let $L$ be the line parallel to the vector $\sqrt{2} \hat{i}-5 \hat{j}+3 \hat{k}$ and passing through the point $A$ given by $\hat{i}+2 \hat{j}-3 \hat{k}$. If the distance between $A$ and a point $P$ on the line $L$ is $18$ units, then the position vector of such a point $P$ is

Let $L_1: \frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}$ and $L_2: \frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{\alpha}, \alpha \in R$,be two lines,which intersect at the point $B$. If $P$ is the foot of perpendicular from the point $A(1,1,-1)$ on $L_2$,then the value of $26 \alpha(PB)^2$ is . . . . . . .

Let in a $\triangle ABC$,the length of the side $AC$ be $6$,the vertex $B$ be $(1,2,3)$ and the vertices $A, C$ lie on the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$. Then the area (in sq. units) of $\triangle ABC$ is

$A$ line $L$ passes through points $A(1, 3, 2)$ and $B(2, 2, 1)$. If the mirror image of point $P(1, 1, -1)$ in the line $L$ is $(x, y, z)$,then $x+y+z=$