⇒Line and Plane

Correct answer option is (A)

Given lines are: $\frac{x+1}{3}=\frac{y+2}{1}=\frac{z+1}{2}$ and $\frac{x-2}{1}=\frac{y+2}{2}=\frac{z-3}{0}$

$\therefore$ Required distance

$=\left|\frac{\left|\begin{array}{lll}3& 0& 4\\ 3& 1& 2\\ 1& 2& 0\end{array}\right|}{\sqrt{(6-1{)}^2+(0-2{)}^2+(0-4{)}^2}}\right|$

$\begin{array}{rl}& =\left|\frac{3(0-4)+0+4(6-1)}{\sqrt{25+4+16}}\right|\\ & =\left|\frac{8}{\sqrt{45}}\right|\\ & =\frac{8}{3\sqrt{5}}\end{array}$

Correct answer option is (D)

The lines arě

$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\text{ and }\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$

Comparing equations with

$\frac{x-x_1}{{\mathrm{a}}_1}=\frac{y-y_1}{{\mathrm{b}}_1}=\frac{\mathrm{z}-{\mathrm{z}}_1}{{\mathrm{c}}_1}\text{ and }\frac{x-x_2}{{\mathrm{a}}_2}=\frac{y-y_2}{{\mathrm{b}}_2}=\frac{\mathrm{z}-{\mathrm{z}}_2}{{\mathrm{c}}_2}\text{, }$

$\begin{array}{rl}& \text{ we get }\\ & x_1=1,y_1=2,{\mathrm{z}}_1=3x_2=2,y_2=4,{\mathrm{z}}_2=5\\ & {\mathrm{a}}_1=2,{\mathrm{b}}_1=3,{\mathrm{c}}_1=4{\mathrm{a}}_2=3,{\mathrm{b}}_2=4,{\mathrm{c}}_2=5\\ & \left|\begin{array}{ccc}{x}_2-x_1\cdot & y_2-y_1& {\mathrm{z}}_2-{\mathrm{z}}_1\\ {\mathrm{a}}_1& {\mathrm{b}}_1& {\mathrm{c}}_1\\ {\mathrm{a}}_2& {\mathrm{b}}_2& {\mathrm{c}}_2\end{array}\right|=\left|\begin{array}{ccc}1& 2& 2\\ 2& 3& 4\\ 3& 4& 5\end{array}\right|\\ & =1(15-16)-2(10-12)+2(8-9)\\ & =1\\ & \therefore \sqrt{{(a_1{b}_2-a_2{b}_1)}^2+{(b_1{c}_2-b_2{c}_1)}^2+{(c_1{a}_2-c_2{a}_1)}^2}\\ & =\sqrt{(2\times 4-3\times 3{)}^2+(3\times 5-4\times 4{)}^2+(4\times 3-5\times 2{)}^2}\\ & =\sqrt{1+1+4}\\ & =\sqrt{6}\end{array}$

Shortest distance between line is $\mathrm{d}$

$\begin{array}{rl}\mathrm{d}& =\frac{\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|}{\sqrt{{(a_1{b}_2-a_2{b}_1)}^2+{(b_1{c}_2-b_2{c}_1)}^2+\sqrt{{(c_1{a}_2-c_2{a}_1)}^2}}}\\ & =\frac{1}{\sqrt{6}}\end{array}$