Finding the area of $\triangle ABC$ with $B$ and $C$ lying on the line $\frac{x-2}{3}=\frac{y–1}{0}=\frac{z}{4}$ 
The vertices $B$ and $C$ of a $\triangle ABC$ lie on the line, $\frac{x-2}{3}=\frac{y–1}{0}=\frac{z}{4}$ such that $BC = 5$ units. Then the area (in sq. units) of this triangle, given that the point $A(1, –1, 2)$, is: $5 \sqrt{17} /\sqrt{34}/ 6 / 2\sqrt{34}?$

My attempt: Let $B$ be $(3k+2,1,4k)$ and $C$ be $(3l+2,1,4l)$. So, using distance formula between $B$ and $C$, I get $k-l=1$. Now, using determinant, where the first row is coordinates of $A$, the second is of $B$, and the third $C$, I get area to be $5$. But the answer is given as $\sqrt{34}$. What is wrong in my method?

Let $AD$ be an altitude of the triangle.
Thus, $$D(2+3t,1,4t)$$ and since $$\vec{AD}\cdot\vec{(3,0,4)}=0,$$ we obtain:
$$(1+3t,2,4t-2)(3,0,4)=0,$$ which gives $$t=\frac{1}{5}.$$
Can you end it now?
