The lines $\dfrac{x-1}{2}=\dfrac{y-4}{4}=\dfrac{z-2}{3}$ and $\dfrac{1-x}{1}=\dfrac{y-2}{5}=\dfrac{3-z}{a}$ are perpendicular to each other, then $a$ equals to

The first line is given by $\dfrac{x-1}{2} = \dfrac{y-4}{4} = \dfrac{z-2}{3}$. The direction vector of this line is $\vec{v_1} = 2\hat{i} + 4\hat{j} + 3\hat{k}$.

The second line is given by $\dfrac{1-x}{1} = \dfrac{y-2}{5} = \dfrac{3-z}{a}$. Rewriting this in standard form $\dfrac{x-1}{-1} = \dfrac{y-2}{5} = \dfrac{z-3}{-a}$, the direction vector is $\vec{v_2} = -1\hat{i} + 5\hat{j} - a\hat{k}$.

Since the lines are perpendicular, the dot product of their direction vectors must be zero, i.e., $\vec{v_1} \cdot \vec{v_2} = 0$.

$(2)(-1) + (4)(5) + (3)(-a) = 0$

$-2 + 20 - 3a = 0$

$18 - 3a = 0$

$3a = 18$

$a = 6$

Answer: 6