The direction ratios of the line of intersection of the planes 3x + 2y – z = 5 and x – y + 2z = 3 are

The correct option is -3, 7, 5.

Step-by-Step Explanation:

Let the direction ratios of the line of intersection of the two planes be represented by:
$a,b,c$

The equations of the two given planes are:
1) $3x+2y-z=5$
2) $x-y+2z=3$

The normal vector to the first plane has direction ratios:
$(3,2,-1)$

The normal vector to the second plane has direction ratios:
$(1,-1,2)$

Since the line of intersection lies in both planes, it is perpendicular to the normal vectors of both planes. Therefore, the dot product of the direction ratios of the line and the normal vectors of the planes must be zero. This gives us the following system of equations:
$3a+2b-c=0$
$a-b+2c=0$

We can solve for the ratio of $a$, $b$, and $c$ using the rule of cross-multiplication:
$\frac{a}{\left(2\right)\left(2\right)-(-1)(-1)}=\frac{-b}{\left(3\right)\left(2\right)-(-1)\left(1\right)}=\frac{c}{\left(3\right)(-1)-\left(2\right)\left(1\right)}$

Simplifying the denominators:
For $a$: $4-1=3$
For $-b$: $6-(-1)=7$, which implies $\frac{-b}{7}=\frac{b}{-7}$
For $c$: $-3-2=-5$

Substituting these values back into the ratios:
$\frac{a}{3}=\frac{b}{-7}=\frac{c}{-5}$

Thus, the direction ratios are proportional to:
$(3,-7,-5)$

Multiplying each component by $-1$ to match the options:
$(-3,7,5)$

Therefore, the direction ratios of the line of intersection are -3, 7, 5.