NDA 2026 — Mathematics PYQ

The distance $d$ between two points $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ is given by:

$$d^2=(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2$$

Step 1: Calculate the squares of the distances

• For $P{Q}^2$: Points $P(-2,-3,5)$ and $Q(4,-1,5)$

  $$P{Q}^2=(4-(-2){)}^2+(-1-(-3){)}^2+(5-5{)}^2$$

  $$P{Q}^2=(6{)}^2+(2{)}^2+(0{)}^2=36+4+0=40$$

• For $Q{S}^2$: Points $Q(4,-1,5)$ and $S(2,-6,10)$

  $$Q{S}^2=(2-4{)}^2+(-6-(-1){)}^2+(10-5{)}^2$$

  $$Q{S}^2=(-2{)}^2+(-5{)}^2+(5{)}^2=4+25+25=54$$

• For $P{R}^2$: Points $P(-2,-3,5)$ and $R(6,-4,8)$

  $$P{R}^2=(6-(-2){)}^2+(-4-(-3){)}^2+(8-5{)}^2$$

  $$P{R}^2=(8{)}^2+(-1{)}^2+(3{)}^2=64+1+9=74$$

Step 2: Substitute values into the expression

We need to find the value of $P{Q}^2+2Q{S}^2-2P{R}^2$:

$$=40+2(54)-2(74)$$

$$=40+108-148$$

$$=148-148$$

$$=0$$

Correct Option: (b) $0$