⇒Line and Plane

Correct answer option is (D)

$\begin{array}{rl}& \bar{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}})+\lambda (\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})+\mu (\hat{\mathrm{i}}-2\hat{\mathrm{j}}+3\hat{k})\\ & \bar{\mathrm{a}}=\hat{\mathrm{i}}-\hat{\mathrm{j}},\bar{\mathrm{A}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}},\bar{\mathrm{B}}=\hat{\mathrm{i}}-2\hat{j}+3\hat{k}\end{array}$

$\overline{n}$ is $\mathrm{\perp }$er to $\bar{\mathrm{A}}$ and $\bar{\mathrm{B}}$

$\begin{array}{rl}& \begin{array}{rl}\therefore \bar{\mathrm{n}}=\bar{\mathrm{A}}\times \bar{\mathrm{B}}& =\left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ 1& 1& 1\\ 1& -2& 3\end{array}\right|\\ & =\hat{\mathrm{i}}(3+2)-\hat{\mathrm{j}}(3-1)+\hat{\mathrm{k}}(-2-1)=5\hat{\mathrm{i}}-2\hat{\mathrm{j}}-3\hat{\mathrm{k}}\end{array}\\ & \begin{array}{c}\bar{\mathrm{a}}\cdot \bar{\mathrm{n}}=\mathrm{d}=(\hat{\mathrm{i}}-\hat{\mathrm{j}})\cdot (5\hat{\mathrm{i}}-2\hat{\mathrm{j}}-3\hat{\mathrm{k}})=5+2=7\end{array}\\ & \begin{array}{c}\therefore \bar{\mathrm{r}}\cdot (5\hat{\mathrm{i}}-2\hat{\mathrm{j}}-3\hat{\mathrm{k}})=7\Rightarrow \text{ Cartesian equation is }5\mathrm{x}-2\mathrm{y}-3\mathrm{z}-7=0\end{array}\end{array}$

Correct answer option is (B)

The equation of the required plane is $(x+2y+3z-4)+\lambda (2x+y-z+5)=0$ i.e. ...... (1)

$(1+2\lambda )x+(2+\lambda )y+(3-\lambda )z+(-4+5\lambda )=0$

Since (1) is perpendicular to the plane $5x+3y-6z+8=0$, we write

$\begin{array}{rl}& (1+2\lambda )(5)+(2+\lambda )(3)+(3-\lambda )(-6)=0\\ & \therefore 5+10\lambda +6+3\lambda -18+6\lambda =0\end{array}$

$19\lambda =7\Rightarrow \lambda =\frac{7}{19}$

Substituting value of $\lambda$ in eq. (1), we get

$\begin{array}{rl}& (1+\frac{14}{19})x+(2+\frac{7}{19})y+(3-\frac{7}{19})z+(-4+\frac{35}{19})=0\\ & \therefore 33x+45y+50z-41=0\end{array}$

Correct answer option is (C)

The plane passes through the point $(0,7,-7)$ and contains the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$.

$\therefore$ Required equation of the plane is

$\begin{array}{rl}& \left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ -1-0& 3-7& -2+7\\ -3& 2& 1\end{array}\right|=0\Rightarrow \left|\begin{array}{ccc}\hat{\mathrm{i}}& \hat{\mathrm{j}}& \hat{\mathrm{k}}\\ -1& -4& 5\\ -3& 2& 1\end{array}\right|=0\\ & \therefore \hat{\mathrm{i}}(-4-10)-\hat{\mathrm{j}}(-1+15)+\hat{\mathrm{k}}(-2-12)=0\\ & \therefore -14\hat{\mathrm{i}}-14\hat{\mathrm{j}}-14\hat{\mathrm{k}}=0\Rightarrow \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}=0\end{array}$

Hence Cartesian equation of the plane is $x+y+z=0$