The lines $\frac{x-2}{2}=\frac{y}{-2}=\frac{z-7}{16}$ and $\frac{x+3}{4}=\frac{y+2}{3}=\frac{z+2}{1}$ intersect at the point $P$. If the distance of P from the line $\frac{x+1}{2}= \frac{\mathrm{y}-1}{3}=\frac{\mathrm{z}-1}{1}$ is $l$, then $14 l^2$ is equal to.
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