Let → a = 2 ^ i + ^ j + ^ k , and → b and → c be two nonzero...

Given,

$\left|\overset{\to}{a} + \overset{\to}{b} + \overset{\to}{c}\right| = \left|\overset{\to}{a} + \overset{\to}{b} - \overset{\to}{c}\right|</p><p>Now squaring both side we get,</p><p>$⇒ {\left|\overset{\to}{a} + \overset{\to}{b} + \overset{\to}{c}\right|}^{2} = {\left|\overset{\to}{a} + \overset{\to}{b} - \overset{\to}{c}\right|}^{2}$</p><p>$⇒ {\left|\overset{\to}{a}\right|}^{2} + {\left|\overset{\to}{b}\right|}^{2} + {\left|\overset{\to}{c}\right|}^{2} + 2 \left(\overset{\to}{a} \cdot \overset{\to}{b}\right) + 2 \left(\overset{\to}{b} \cdot \overset{\to}{c}\right) + 2 \left(\overset{\to}{c} \cdot \overset{\to}{a}\right) = {\left|\overset{\to}{a}\right|}^{2} + {\left|\overset{\to}{b}\right|}^{2} + {\left|\overset{\to}{c}\right|}^{2} + 2 \left(\overset{\to}{a} \cdot \overset{\to}{b}\right) - 2\left(\overset{\to}{b} \cdot \overset{\to}{c}\right) - 2\left(\overset{\to}{c} \cdot \overset{\to}{a}\right)</p><p>⇒ 2 \left(\overset{\to}{b} \cdot \overset{\to}{c}\right)$+ 2 \left(\overset{\to}{c} \cdot \overset{\to}{a}\right)$ = - 2 \left(\overset{\to}{b} \cdot \overset{\to}{c}\right) + 2 \left(\overset{\to}{c} \cdot \overset{\to}{a}\right)

$⇒ 4\left[\left(\overset{\to}{b} \cdot \overset{\to}{c}\right) + \left(\overset{\to}{c} \cdot \overset{\to}{a}\right)\right] = 0$

Now here given $\overset{\to}{b} \cdot \overset{\to}{c} = 0$, so

$\overset{\to}{c} \cdot \overset{\to}{a} = 0$ which means $\overset{\to}{c} \text{\text{ and } } \overset{\to}{a}$ are perpendicular vector, so statement \left(B\right)$is false,</p><p>Now statement \left(A\right)$: \left|\overset{\to}{a} + \lambda\overset{\to}{c}\right| \ge \left|\overset{\to}{a}\right|

$⇒ {\left|\overset{\to}{a} + \lambda\overset{\to}{c}\right|}$^{2} \ge {\left|\overset{\to}{a}\right|}^{2}

$⇒ {\left|\overset{\to}{a}\right|}$^{2} + \lambda{\left|\overset{\to}{c}\right|}^{2} + 2\left(\overset{\to}{a} \cdot \overset{\to}{c}\right) \ge {\left|\overset{\to}{a}\right|}^{2}

$⇒ \lambda{\left|\overset{\to}{c}\right|}$^{2} + 2\left(\overset{\to}{a} \cdot \overset{\to}{c}\right)$\ge 0$

$⇒ \lambda{\left|\overset{\to}{c}\right|}$^{2} \ge 0$</p><p>This is not always true, since$\lambda \in R$.</p><p>Hence, we can say that statement \left(A\right) is true$ and $statement \left(B\right) is false.</p>$