Explanation:
C 
Let co-ordinate of \(\mathrm{D}\) is \((\mathrm{x}, \mathrm{y}, \mathrm{z})\).
Using mid-point formula,
\(\mathrm{Q}=\left(\frac{\mathrm{x}-6}{2}, \frac{\mathrm{y}-9}{2}, \frac{\mathrm{z}+0}{2}\right)\)
Also, \(\mathrm{P}=\left(\frac{2+6}{2}, \frac{3+9}{2}, \frac{0+0}{2}\right)=(4,6,0)\)
Since, \(A C \| P Q\)
\(\therefore\) D.r's of line \(A C=D\).r's of line \(P Q\)
\(\Rightarrow(-8,-12,0)=\left(\frac{x-14}{2}, \frac{y-21}{2}, \frac{z}{2}\right)\)
\(\Rightarrow \mathrm{x}=-2, \mathrm{y}=-3, \mathrm{z}=0\)
\(\Rightarrow \mathrm{D}(-2,-3,0)=\mathrm{Q}(-4,-6,0)\)
If \(L\) is midpoint of \(P Q\), then
\(\mathrm{L}=\left(\frac{4-4}{2}, \frac{6-6}{2}, 0\right)=(0,0,0)\)
\(\therefore \perp \text { distance of } \mathrm{L}(0,0,0) \text { from the plat }\)
\(3 \mathrm{x}+4 \mathrm{z}+25=0 \text { is }\)
\(\mathrm{a}_1=0, \quad \mathrm{~b}_1=0, \quad \mathrm{c}_1=0\)
\(\mathrm{a}_2=3, \quad \mathrm{~b}_2=0, \quad \mathrm{c}_2=4, \quad \mathrm{~d}_2=25\)
\(\mathrm{~d}=\frac{\mathrm{a}_1 \mathrm{a}_2+\mathrm{b}_1 \mathrm{~b}_2+\mathrm{c}_1 \mathrm{c}_2+\mathrm{d}_2}{\sqrt{\mathrm{a}_2^2+\mathrm{b}_2^2+\mathrm{c}_2^2}}\)
\(=\frac{0 \times 3+0 \times 0+0 \times 4+25}{\sqrt{(3)^2+(0)^2+(4)^2}}\)
\(\Rightarrow \mathrm{d}=\frac{25}{5}=5\)\(\therefore \perp\) distance of \(L(0,0,0)\) from the plane