⇒Line and Plane

Correct answer option is (B)

We have lines $y-2=0$ and $4x-3z+5=0$

$\begin{array}{rl}& \therefore 4x=3z-5=3[z-\left(\frac{5}{3}\right)]\\ & \therefore \frac{4x}{12}=\frac{3[z-\left(\frac{5}{3}\right)]}{12}\Rightarrow \frac{x}{3}=\frac{z-\left(\frac{5}{3}\right)}{4},y=2\end{array}$

Thus line passes through the point $(0,2,\frac{5}{3})$ i.e. a point having position vector $2\mathrm{j}+\frac{5}{3}\hat{k}$

Also direction ratios of a line are $3,0,4$.

Hence required vector equation is

$\bar{\mathrm{r}}=(2\hat{\mathrm{j}}+\frac{5}{3}\hat{\mathrm{k}})+\lambda (3\hat{\mathrm{i}}+4\hat{\mathrm{k}})$

Correct answer option is (D)

The direction ratios of the line passing through the points A(3,4,-7) and B(1,-1,6) can be found by subtracting the coordinates of A from B :

$\text{Direction ratio of x (l)}=1-3=-2$

$\text{Direction ratio of y (m)}=-1-4=-5$

$\text{Direction ratio of z (n)}=6+7=13$

So, the direction ratios of the line are $(-2,-5,13)$.

The parametric equation of a line passing through $(x_1,y_1,z_1)$ and having direction ratios $(l,m,n)$ is :

$\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}=\lambda$

Where $\lambda$ is a parameter.

Using the above formula and the coordinates of point A and the direction ratios :

$\frac{x-3}{-2}=\frac{y-4}{-5}=\frac{z+7}{13}=\lambda$

From this equation, the parametric equations are :

$x=3-2\lambda$

$y=4-5\lambda$

$z=-7+13\lambda$

Thus, the correct option is :

Option D :

$x=3-2\lambda ,y=4-5\lambda ,z=-7+13\lambda$