If the $x$-coordinate of a point $P$ on the line joining the points $Q(2, 2, 1)$ and $R(5, 2, -2)$ is $4$,then the $y$-coordinate of $P$ is:

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

If the lines $\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-1}{4}$ and $\frac{x-3}{1} = \frac{y-k}{2} = \frac{z}{1}$ intersect,then find the value of $k$.

The square of the distance of the point of intersection of the lines $\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(\hat{i} - \hat{j})$ and $\vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + \hat{k})$ from the origin is:

The length of the perpendicular drawn from the origin to the line $\bar{r} = (4\hat{i} + 2\hat{j} + 4\hat{k}) + \lambda(3\hat{i} + 4\hat{j} - 5\hat{k})$ is .......

Let a triangle $PQR$ be such that $P$ and $Q$ lie on the line $\frac{x+3}{8} = \frac{y-4}{2} = \frac{z+1}{2}$ and are at a distance of $6$ units from $R(1, 2, 3)$. If $(\alpha, \beta, \gamma)$ is the centroid of $\triangle PQR$,then $\alpha + \beta + \gamma$ is equal to :