Given equation of line $\frac{x-1}{3}=\frac{y+2}{4}=\frac{z-3}{-2}$ and equation of a place $2x-y +3z=1$. Which of the following is correct ?

A. Given line and plance intersect ar (10,10, -3).

B. The line passes through the point (-1, 2, -3)

C. Direction Ratio's of normal to the plane are -2, -1, -3.

D. The line is parallel to the vector $3\hat{i}+4\hat{j}-2\hat{k}$.

Choose the correct answer from the options given below.

A and D only

The correct answer is option (2) → A and D only

$\frac{x-1}{3}=\frac{y+2}{4}=\frac{z-3}{-2}=λ$

line passes through (1, -2, 3) parallel to $3\hat i+4\hat j-2\hat k$

so $x=3λ+1,y=4λ-2,z=-2λ+3$

plane : $2x-y+3z=1$

$\vec n$ ⊥ plane → $2\hat i-\hat j+3\hat k$

so finding intersection

$2(3λ+1)-(4λ-2)+3(-2λ+3)=1$

$⇒6λ-4λ-6λ+2+2+9=1$

so $12=4λ⇒λ=3$ 

point of intersection → $x=10,y=10,z=-3$

Only A, D correct