1.Let a⃗=a1i^+a2j^+a3k^\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}a=a1i^+a2j^+a3k^, b⃗=b1i^+b2j^+b3k^\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}b=b1i^+b2j^+b3k^, c⃗=c1i^+c2j^+c3k^\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}c=c1i^+c2j^+c3k^ be three non-zero vectors such that c⃗\vec{c}c is a unit vector perpendicular to both a⃗\vec{a}a and b⃗\vec{b}b. If a⃗⊥b⃗\vec{a} \perp \vec{b}a⊥b, then ∣[a⃗ b⃗ c⃗]∣2=|[\vec{a}\;\vec{b}\;\vec{c}]|^2 =∣[abc]∣2=a.000b.111c.(a12+a22+a32)(b12+b22+b32)(a_1^2 + a_2^2 + a_3^2)(b_1^2 + b_2^2 + b_3^2)(a12+a22+a32)(b12+b22+b32)d.(a12+a22+a32)(b12+b22+b32)(c12+c22+c32)(a_1^2 + a_2^2 + a_3^2)(b_1^2 + b_2^2 + b_3^2)(c_1^2 + c_2^2 + c_3^2)(a12+a22+a32)(b12+b22+b32)(c12+c22+c32)Login to continueOnly logged in users canattempt or see the solution.