$P$ is a point on the line segment joining the points $(3,2,-1)$ and $(6,2,-2)$. If $x$ coordinate of $\mathrm{P}$ is 5, then its $y$ co-ordinate is

Let the points be $A(3, 2, -1)$ and $B(6, 2, -2)$. The line segment joining $A$ and $B$ can be represented by the parametric equations $x = 3 + (6-3)t = 3 + 3t$, $y = 2 + (2-2)t = 2$, and $z = -1 + (-2 - (-1))t = -1 - t$, where $0 \le t \le 1$.

Given that the $x$-coordinate of point $P$ is $5$, we set $3 + 3t = 5$.

Solving for $t$, we get $3t = 2$, which implies $t = \dfrac{2}{3}$.

Since $t = \dfrac{2}{3}$ lies in the interval $[0, 1]$, the point $P$ lies on the segment $AB$.

The $y$-coordinate of any point on this line segment is given by $y = 2 + (2-2)t = 2$.

Therefore, the $y$-coordinate of point $P$ is $2$.

Answer: 2