Q.Find the length of the perpendicular from $(2, -3, 1)$ to the line $\dfrac{x+1}{2} = \dfrac{y-3}{3} = \dfrac{z+1}{-1}$

[!TLDR] Drop a perpendicular from $(2,-3,1)$ to the line; the foot is the point on the line whose join to $(2,-3,1)$ is perpendicular to the line's direction.

Taking a general point $(2\lambda-1, 3\lambda+3, -\lambda-1)$ on the line and requiring its join to $(2,-3,1)$ to be perpendicular to direction $(2,3,-1)$ gives $\lambda=-1$, so the foot is $(-3,0,0)$ and the perpendicular length is $\sqrt{35}$.

[!ANSWER] Length of the perpendicular $=\sqrt{35}$ unit.

The given line is $\dfrac{x+1}{2}=\dfrac{y-3}{3}=\dfrac{z+1}{-1}=\lambda$ (say), through $A(-1,3,-1)$ with direction ratios $(2,3,-1)$.

A general point on the line is $M(2\lambda-1,\ 3\lambda+3,\ -\lambda-1)$.

Let $P(2,-3,1)$. The direction ratios of $PM$ are

Since $PM$ is the required perpendicular, $PM\perp$ line, so $PM\cdot(2,3,-1)=0$:

So $M=(2(-1)-1,\ 3(-1)+3,\ -(-1)-1)=(-3,\ 0,\ 0)$.

[!ANSWER] Length of the perpendicular from $(2,-3,1)$ to the given line $=\sqrt{35}$ unit.

Take a general point on the line using the parameter $\lambda$, form the direction ratios of the join to the given point, impose perpendicularity with the line's direction ratios to solve for $\lambda$ and locate the foot, then use the distance formula.

Writing the general point on the line incorrectly (forgetting the line passes through $(-1,3,-1)$, not the origin); imposing the perpendicularity dot product with the wrong sign convention for $PM$; arithmetic slips while expanding $14\lambda+14=0$.